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JEEMAIN.GURU, , JEE-Mathematics, , CIRCLE, 1. (A) DEFINITION :, A circle is the locus of a point which moves in a plane in such a way that its distance from a fixed point (in the, same given plane) remains constant. The fixed point is called the centre of the circle and the constant distance, is called the radius of the circle., Equation of a circle :, The curve traced by the moving point is called its circumference i.e. the equation of any circle is satisfied by, co-ordinates of all points on its circumference., or, The equation of the circle means the equation of its circumference., or, It is the set of all points lying on the circumference of the circle., Chord and diameter - the line joining any two points on the circumference is called a, P, Q, C, chord. If any chord passing through its centre is called its diameter., A, , AB = chord, PQ = diameter, , B, , (B) BASIC THEOREMS & RESULTS OF CIRCLES :, (a), Concentric circles : Circles having same centre., (b), Congruent circles : Iff their radii are equal., (c), Congruent arcs : Iff they have same degree measure at the centre., Theorem 1 :, (i), If two arcs of a circle (or of congruent circles) are congruent, the corresponding chords are equal., Converse : If two chords of a circle are equal then their corresponding arcs are congruent., (ii), Equal chords of a circle (or of congruent circles) subtend equal angles at the centre., Converse : If the angle subtended by two chords of a circle (or of congruent circles) at the centre, are equal, the chords are equal., Theorem 2 :, (i), The perpendicular from the centre of a circle to a chord bisects the chord., Converse : The line joining the mid point of a chord to the centre of a circle is perpendicular to, the chord., (ii), Perpendicular bisectors of two chords of a circle intersect at its centre., Theorem 3 :, (i), There is one and only one circle passing through three non collinear points., (ii), If two circles intersects in two points, then the line joining the centres is perpendicular bisector of, common chords., Theorem 4 :, (i), Equal chords of a circle (or of congruent circles) are equidistant from the centre., Converse : Chords of a circle (or of congruent circles) which are equidistant from the centre are, equal in length., (ii), If two equal chords are drawn from a point on the circle, then the centre of circle will lie on angle, bisector of these two chords., (iii), Of any two chords of a circle larger will be near to centre., , Theorem 5 :, O, (i), The degree measure of an arc or angle subtended by an arc at the centre is, 2, double the angle subtended by it at any point of alternate segment., , (ii), , Angle in the same segment of a circle are equal., , 44, , , , , , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , C = centre, , E

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JEEMAIN.GURU, (iii), , JEE-Mathematics, , The angle in a semi circle is right angle., Converse : The arc of a circle subtending a right angle in alternate segment, is semi circle., , Theorem 6 :, Any angle subtended by a minor arc in the alternate segment is acute and any angle subtended by a, major arc in the alternate segment is obtuse., Theorem 7 :, If a line segment joining two points subtends equal angles at two other points lying on the same side of the, line segment, the four points are concyclic, i.e. lie on the same circle., , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , (d), , E, , Cyclic Quadrilaterals :, A quadrilateral is called a cyclic quadrilateral if its all vertices lie on a circle., Theorem 1 :, The sum of either pair of opposite angles of a cyclic quadrilateral is 180°, OR, The opposite angles of a cyclic quadrilateral are supplementary., Converse : If the sum of any pair of opposite angle of a quadrilateral is 180°, then the quadrilateral is, cyclic., Theorem 2 :, If a side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite, angle., Theorem 3 :, C, D, The internal angle bisectors of a cyclic quadrilateral form a quadrilateral which is, P, Q, S, also cyclic., R, Theorem 4 :, A, B, If two sides of a cyclic quadrilateral are parallel then the remaining two sides are, equal and the diagonals are also equal., OR, A cyclic trapezium is isosceles and its diagonals are equal., Converse : If two non-parallel sides of a trapezium are equal, then it is cyclic., OR, An isosceles trapezium is always cyclic., Theorem 5 :, When the opposite sides of cyclic quadrilateral (provided that they are not parallel) are produced, then, the exterior angle bisectors intersect at right angle., , (C) TANGENTS TO A CIRCLE :, Theorem 1 :, A tangent to a circle is perpendicular to the radius through the point of contact., Converse : A line drawn through the end point of a radius and perpendicular to it is a tangent to the circle., Theorem 2 :, If two tangents are drawn to a circle from an external point, then :, , , (i), they are equal., , , (ii), they subtend equal angles at the centre,, (iii), they are equally inclined to the segment, joining the centre to that point., Theorem 3 :, A, D, A, If two chords of a circle intersect inside or outside the circle, B, O, when produced, the rectangle formed by the two segments, P, of one chord is equal in area to the rectangle formed by the, P, D, two segments of the other chord., C, B, C, PA × PB = PC × PD, , 45

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JEEMAIN.GURU, , JEE-Mathematics, Theorem 4 :, If PAB is a secant to a circle intersecting the circle at A and B and PT is, tangent segment, then PA × PB = PT2, OR, Area of the rectangle formed by the two segments of a chord is equal to, the area of the square of side equal to the length of the tangent from the, point on the circle., , B, A, O, , P, , T, , E, , Theorem 5 :, , C, , If a chord is drawn through the point of contact of a tangent to a circle,, B, O, then the angles which this chord makes with the given tangent are equal, respectively to the angles formed in the corresponding alternate segments., D, BAQ = ACB and BAP = ADB, A, P, Q, Converse :, If a line is drawn through an end point of a chord of a circle so that the angle formed with the chord is equal to, the angle subtended by the chord in the alternate segment, then the line is a tangent to the circle., STANDARD EQUATIONS OF THE CIRCLE :, (a), , Central Form :, If (h, k) is the centre and r is the radius of the circle then its equation is, (x–h) 2 + (y–k) 2 = r 2, Special Cases :, (i), (ii), (iii), , If centre is origin (0,0) and radius is 'r' then equation of circle is x2 + y 2 = r 2, and this is called the standard form., If radius of circle is zero then equation of circle is (x – h)2 + (y – k)2 = 0., y, Such circle is called zero circle or point circle., When circle touches x-axis then equation of the circle is, (h,k), , (x–h) 2 + (y–k) 2 = k 2 ., , C, k, 0 Touching x-axis, , (iv), , When circle touches y-axis then equation of circle is, (x–h) 2 + (y–k) 2 = h 2, , y, C (h,k), h, , ., , 0 Touching y-axis, , (v), , x, , When circle touches both the axes (x-axis and y-axis) then equation of, , x, , y, , circle (x–h) 2 + (y–h) 2 = h 2 ., C(h,h), h, h, 0 Touching x-axis, , x, , and y-axis, , (vi), , When circle passes through the origin and centre of the circle is (h,k), then radius, , h 2  k 2  r and intercept cut on x-axis OP =2h,, , y, , Q, , (0,2k), , and intercept cut on y-axis is OQ = 2k and equation of circle is, (x–h)2 + (y–k)2 = h2 + k2 or x2 + y2 – 2hx – 2ky = 0, , C(h,k), k, , O, , (2h,0), , P, , x, , Note : Centre of the circle may exist in any quadrant hence for general cases use ± sign before h & k., , 46, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , 2., , E

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JEEMAIN.GURU, (b), , JEE-Mathematics, , General equation of circle, x2 + y 2 + 2gx + 2fy + c = 0. where g,f,c are constants and centre is (–g,–f), ,  coefficient of x coefficient of y , 2, 2, ,, i.e.  ,  and radius r  g  f  c, , 2, 2, Note, (i), (ii), (iii), (iv), , :, If (g2 + f2 – c) > 0, then r is real and positive and the circle is a real circle., If (g2 + f2 – c) = 0, then radius r = 0 and circle is a point circle., If (g2 + f2 –c)<0, then r is imaginary then circle is also an imaginary circle with real centre., x2 + y2 + 2gx + 2fy + c = 0, has three constants and to get the equation of the circle at least three, conditions should be known  A unique circle passes through three non collinear points., The general second degree in x and y, ax 2 + by 2 + 2hxy + 2gx + 2fy + c = 0 represents, a circle if :, , (v), , •, •, •, (c), , coefficient of x2 = coefficient of y2 or a = b  0, coefficient of xy = 0 or h = 0, (g2 + f2 – c)  0 (for a real circle), , Intercepts cut by the circle on axes :, The intercepts cut by the circle x2 + y2 + 2gx + 2fy + c =0 on :, x-axis = 2 g 2 – c, , (i), Note, (i), (ii), (iii), (iv), (v), (vi), (vii), , (ii), , y-axis = 2 f 2 – c, , :, If the circle cuts the x-axis at two distinct point, then g2 – c > 0, If the cirlce cuts the y-axis at two distinct point, then f2 – c > 0, If circle touches x-axis then g2 = c., If circle touches y-axis then f2 = c., Circle lies completely above or below the x-axis then g2 < c., Circle lies completely to the right or left to the y-axis, then f2 < c., Intercept cut by a line on the circle x2 + y 2 + 2gx + 2fy+c=0 or length of, , a, A, , chord of the circle  2 a 2  P 2 where a is the radius and P is the length of, , O, P, C, , B, , perpendicular from the centre to the chord., , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , (d), , E, , Equation of circle in diameter form :, , P(x,y), , If A(x1,y1) and B(x2,y2) are the end points of the diameter of the circle, and P(x,y) is the point other then A and B on the circle then from, geometry we know that APB = 90°., , (Slope of PA) × (Slope of PB) = –1, , , (e), , (x1,y1) A, , C, , B(x2,y2), ,  y  y1   y  y 2 , , = – 1,  x  x 1   x  x 2 , (x–x 1 ) (x–x 2 )+(y–y 1 )(y–y 2 ) = 0, , Note : This will be the circle of least radius passing through (x1, y1) and (x2, y2), Equation of circle in parametric forms :, (i), The parametric equation of the circle x2+y2 = r2 are x = r cos, y = r sin ;   [0, 2) and, (ii), , (r cos , r sin ) are called the parametric co-ordinates., The parametric equation of the circle (x – h)2 + (y – k)2 = r2 is x = h + r cos, y = k + r sin , where  is parameter., , (iii), , The parametric equation of the circle x2 + y2 + 2gx + 2fy + c = 0 are x = – g +, y = –f +, , g 2 + f 2 – c sin  where  is parameter.., , 47, , g 2 + f 2 – c cos,

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JEEMAIN.GURU, , JEE-Mathematics, Note : Equation of a straight line joining two point    on the circle x2 + y2 = a2 is, x cos, , Illustration 1 :, , Solution :, ,  ,  ,  , + y sin, = a cos, ., 2, 2, 2, , Find the centre and the radius of the circles, (a), 3x2 + 3y2 – 8x – 10y + 3 = 0, (b), x2 + y2 + 2x sin + 2y cos – 8 = 0, (c), 2x2 + xy + 2y2 + ( – 4)x + 6y – 5 = 0, for some ., (a), We rewrite the given equation as, x2 + y2 –, , (b), , 8, 10, x, y 1  0, 3, 3, , , , g = –, , 4, 5, , f = – , c = 1, 3, 3, ,  4 5, Hence the centre is  ,  and the radius is, 3 3, 2, 2, x + y + 2x sin + 2ycos – 8 = 0., Centre of this circle is (–sin, – cos), , 16 25, , 1 , 9, 9, , 32 4 2, , units, 9, 3, , sin 2   cos 2   8  1  8  3 units, 2x 2 + xy + 2y 2 + ( – 4)x + 6y – 5 = 0, We rewrite the equation as, Radius =, , (c), , x2 , , , 5,    4, xy  y 2  ,  x  3y   0, , 2, 2, 2, , ........ (i), , Since, there is no term of xy in the equation of circle , So, equation (i) reduces to, , x 2 + y 2 – 2x + 3y , , , = 0     = 0, 2, , 5, 0, 2, , 3, , 9 5, 23, centre is  1,  , Radius = 1   , units., 2, 4 2, 2, If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, then the radius of the circle, is (A) 3/2, (B) 3/4, (C) 1/10, (D) 1/20, The diameter of the circle is perpendicular distance between the parallel lines (tangents), , , Solution :, , 3x – 4y + 4 = 0 and 3x – 4y –, Hence radius is, , 7, 4 7/2 3,  ., = 0 and so it is equal to, 2, 9  16 2, , 3, ., 4, , Ans. (B), , Illustration 3 :, , If y = 2x + m is a diameter to the circle x2 + y2 + 3x + 4y – 1 = 0, then find m, , Solution :, , Centre of circle = (–3/2 , –2). This lies on diameter y = 2x + m, , , , Illustration 4 :, , Solution :, , – 2 = (–3/2) × 2 + m , , m=1, , The equation of a circle which passes through the point ( 1 , –2) and ( 4 , –3) and whose centre lies, on the line 3x + 4y = 7 is, (A) 15 ( x2 + y2) – 94x + 18y – 55 = 0, , (B) 15 ( x2 + y2) – 94x + 18y + 55 = 0, , (C) 15 ( x2 + y2) + 94x – 18y + 55 = 0, , (D) none of these, , 2, , 2, , Let the circle be x + y + 2gx + 2fy + c = 0, , ..... (i), , Hence, substituting the points, ( 1, –2) and ( 4 , –3) in equation (i), 5 + 2g – 4f + c = 0, , ..... (ii), , 25 + 8g – 6f + c = 0, , ..... (iii), , centre ( –g , –f) lies on line 3x + 4y = 7, , 48, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , Illustration 2 :, , E

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JEEMAIN.GURU, , JEE-Mathematics, , Hence –3g –4f = 7, solving for g, f,c, we get, Here g =, , 47, 9, 55, , f, , c, 15, 15, 15, , Hence the equation is 15 ( x2 + y2 ) –94x + 18y + 55 = 0, , Illustration 5 :, Solution :, , Ans. (B), , A circle has radius equal to 3 units and its centre lies on the line y = x – 1. Find the equation of the, circle if it passes through (7, 3)., Let the centre of the circle be ( ). It lies on the line y = x – 1, ,    – 1. Hence the centre is (  –1)., , The equation of the circle is (x – ) 2 + (y –  + 1) 2 = 9, It passes through (7, 3),     (7 – )2 + (4 – )2 = 9, , 22 – 22 + 56 = 0,  2 – 11 + 28 = 0, , ( – 4)( – 7) = 0, ,  = 4, 7, Hence the required equations are, x 2 + y 2 – 8x – 6y + 16 = 0 and x 2 + y 2 – 14x – 12y + 76 = 0., , Ans., , Do yourself - 1 :, (i), , Find the centre and radius of the circle 2x2 + 2y2 = 3x – 5y + 7, , (ii), , Find the equation of the circle whose centre is the point of intersection of the lines 2x – 3y + 4 = 0 &, 3x + 4y – 5 = 0 and passes through the origin., , (i i i ), , Find the parametric form of the equation of the circle x2 + y2 + px + py = 0, , (iv), , Find the equation of the circle the end points of whose diameter are the centres of the circles, x2 + y2 + 16x – 14y = 1 & x2 + y2 – 4x + 10y = 2, , 3., , POSITION OF A POINT W.R.T CIRCLE :, (a), , Let the circle is x2 + y2 + 2gx + 2fy + c = 0 and the point is (x1,y1) then Point (x1,y1) lies out side the circle or on the circle or inside the circle according as, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, ,  x12 + y12 + 2gx1 +2fy1 + c >, =, < 0 or S1 >, =, < 0, , E, , (b), , The greatest & the least distance of a point A from a circle with centre, C & radius r is AC + r, , 4., , & |AC – r| respectively., , POWER OF A POINT W.R.T. CIRCLE :, Theorem : The power of point P(x1, y1) w.r.t. the circle x2 + y2 + 2gx + 2ƒy + c = 0 is S1, where S1 = x12  y 12  2gx 1  2ƒ y 1  c, Note : If P outside, inside or on the circle then power of point is positive,, negative or zero respectively., , T, , If from a point P(x 1 , y 1), inside or outside the circle, a secant be drawn, intersecting the circle in two points A & B, then PA . PB = constant. The, product PA . PB is called power of point P(x1, y1) w.r.t. the circle, S  x2 + y2 + 2gx + 2ƒy + c = 0, i.e. for number of secants PA.PB = PA1 . PB1, = PA2 . PB2 = ...... = PT2 = S1, , 49, , B, , A, P, , A1, , B1

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JEEMAIN.GURU, , JEE-Mathematics, Illustration 6 :, , If P(2, 8) is an interior point of a circle x2 + y2 – 2x + 4y – p = 0 which neither touches nor, intersects the axes, then set for p is (A) p < –1, , Solution :, , (B) p < – 4, , (C) p > 96, , (D) , , For internal point p(2, 8), 4 + 64 – 4 + 32 – p < 0  p > 96, and x intercept = 2 1  p therefore 1 + p < 0, , , p < –1 and y intercept = 2 4  p, , , , p < –4, , Ans. (D), , Do yourself - 2 :, (i), , Find the position of the points (1, 2) & (6, 0) w.r.t. the circle x2 + y2 – 4x + 2y – 11 = 0, , (ii), , Find the greatest and least distance of a point P(7, 3) from circle x2 + y2 – 8x – 6y + 16 = 0. Also find, the power of point P w.r.t. circle., , 5., , TANGENT LINE OF CIRCLE :, When a straight line meet a circle on two coincident points then it is called the tangent of the circle., (a), , Condition of Tangency :, , (P>r), , The line L = 0 touches the circle S = 0 if P the length of the, , (P=r), (P<r), , perpendicular from the centre to that line and radius of the, , (P=0), , Tangent, Secant, , P, , r, , Diameter, , circle r are equal i.e. P = r., , Illustration 7 :, , Find the range of parameter 'a' for which the variable line y = 2x + a lies between the circles, x2 + y2 – 2x – 2y + 1 = 0 and x2 + y2 – 16x – 2y + 61 = 0 without intersecting or touching either, circle., , Solution :, , The given circles are C1 : (x – 1)2 + (y – 1)2 = 1 and C2 : (x – 8)2 + (y – 1)2 = 4, The line y – 2x – a = 0 will lie between these circle if centre of the circles lie on opposite sides of, the line, i.e. (1 – 2 – a)(1 – 16 – a) < 0  a  (–15, –1), , , , |1 + a| >, , , , a>, , 5 –1, , |1  2  a|, 5, ,  1,, , |1  16  a|, 5, , 2, , 5 , |15 + a| > 2 5, , or a < – 5 – 1, a > 2 5 – 15 or a < –2 5 – 15, , Hence common values of 'a' are (2 5 – 15, – 5 –1)., , Illustration 8 :, , Solution :, , The equation of a circle whose centre is (3, –1) and which cuts off a chord of length 6 on the line, 2x– 5y+ 18 = 0, (A) (x – 3)2 + (y + 1)2 = 38, , (B) (x + 3)2 + (y – 1)2 = 38, , (C) (x – 3)2 + (y + 1)2 =, , (D) none of these, , 38, , Let AB(= 6) be the chord intercepted by the line 2x – 5y + 18 = 0, from the circle and let CD be the perpendicular drawn from centre, (3, –1) to the chord AB., , C(3,-1), , 2.3  5( 1)  18, , i.e., AD = 3, CD =, ,  29, , 22  52, , A, , D, , B, , Therefore, CA2 = 32 + ( 29 )2 = 38, Hence required equation is (x – 3)2 + (y + 1)2 = 38, , 50, , Ans. (A), , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , Line wouldn't touch or intersect the circles if,, , E

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JEEMAIN.GURU, Illustration 9 :, , JEE-Mathematics, , The area of the triangle formed by line joining the origin to the points of intersection(s) of the line, , x 5  2 y  3 5 and circle x2 + y2 = 10 is, (A) 3, , Solution :, , (B) 4, , (C) 5, , (D) 6, , Length of perpendicular from origin to the line x 5  2 y  3 5 is, , OL , , 3 5, , , , 3 5, , ( 5 )2  2 2, , 9, , Q, L, ,  5, , Radius of the given circle =, , O, 10, , 10 = OQ = OP, , 2, 2, PQ = 2QL = 2 OQ  OL  2 10  5  2 5, , Thus area of OPQ =, , (b), , P, , 5 x + 2y = 3 5, , 1, 1,  PQ  OL   2 5  5  5, 2, 2, , Ans. (C), , Equation of the tangent (T = 0) :, (i), (ii), , Tangent at the point (x1,y1) on the circle x2+ y2 = a2 is xx 1 + yy 1 = a2., (1) The tangent at the point (acos t, asin t) on the circle x 2 + y 2 = a 2 is xcos t + ysin t = a, (2), ,  a cos    a sin    , 2 ,, 2 , The point of intersection of the tangents at the points P() and Q() is ,    .,  cos  , cos, 2, 2 , , , (iii), , The equation of tangent at the point (x1,y1) on the circle x2 + y2 + 2gx + 2fy + c = 0 is, xx 1 + yy 1 + g(x + x 1) + f(y + y 1) + c = 0, , (iv), , If line y = mx + c is a straight line touching the circle x2 + y2 = a2, then c = ± a 1  m 2 and contact, , , , am, a, , , points are  , , 1 + m2, 1 + m2 , , (v), , or, ,  a2m, a2 , , , ,  and equation of tangent is, c, c , , , y = mx ± a 1 + m 2, The equation of tangent with slope m of the circle (x – h)2 + (y – k)2 = a2 is, (y – k) = m(x – h) ± a 1 + m 2, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , Note : To get the equation of tangent at the point (x1 y1) on any second degree curve we replace xx1 in, , E, , place of x2, yy1 in place of y2,, , x  x1, y  y1, xy 1  yx 1, in place of x,, in place of y,, in place of xy, 2, 2, 2, , and c in place of c., (c), , Length of tangent ( S 1 ) :, , T, P(x1,y1), , The length of tangent drawn from point (x1,y1) out side the circle, S  x2 + y2 + 2gx + 2fy + c = 0 is,, PT= S 1 = x 12  y 12  2gx 1  2fy 1  c, Note : When we use this formula the coefficient of x2 and y2 must be 1., (d), , Equation of Pair of tangents (SS 1 = T 2 ) :, Let the equation of circle S  x2 + y2 = a2 and P(x1,y1) is any point, , Q, , outside the circle. From the point we can draw two real and distinct, tangent PQ & PR and combine equation of pair of tangents is (x2 + y2 – a2) (x12 + y12 – a2) = (xx1 + yy1 – a2)2, SS 1 = T 2, , 51, , or, , M, , P, , (x1,y1), , (0,0), , R

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JEEMAIN.GURU, , JEE-Mathematics, Illustration 10 :, , Let A be the centre of the circle x2 + y2 – 2x – 4y – 20 = 0 and B(1, 7) and D(4, –2) are points, on the circle then, if tangents be drawn at B and D, which meet at C, then area of quadrilateral, ABCD is (A) 150, , (B) 75, , (C) 75/2, , (D) none of these, , B (1, 7), , Solution :, (1, 2), A, , C, (16, 7), D (4, –2), , Here centre A(1, 2) and Tangent at (1, 7) is, x.1 + y.7 – 1(x + 1) – 2(y + 7) – 20 = 0 or y = 7, , .......... (i), , Tangent at D(4, –2) is 3x – 4y – 20 = 0, , .......... (ii), , Solving (i) and (ii), C is (16, 7), Area ABCD = AB × BC = 5 × 15 = 75 units., , Ans. (B), , Do yourself - 3 :, (i), , Find the equation of tangent to the circle x2 + y2 – 2ax = 0 at the point (a(1 + cos), asin)., , (ii), , Find the equations of tangents to the circle x2 + y2 – 6x + 4y – 12 = 0 which are parallel to the line, 4x – 3y + 6 = 0, , (i i i ), , Find the equation of the tangents to the circle x 2 + y 2 = 4 which are perpendicular to the line, 12x – 5y + 9 = 0. Also find the points of contact., , (iv), 6., , Find the value of 'c' if the line y = c is a tangent to the circle x2 + y2 – 2x + 2y – 2 = 0 at the point (1, 1), , NORMAL OF CIRCLE :, Normal at a point is the straight line which is perpendicular to the tangent at the point of contact., Note : Normal at point of the circle passes through the centre of the circle., Equation of normal at point (x1,y1) of circle x2 + y2 + 2gx + 2fy + c = 0 is, , y +f, (x - x1 ), y– y1 =  1,  x 1 + g , , N (–g, –f), P, , T, (x1,y1), , y y1, , x x1, , (b), , The equation of normal on any point (x1,y1) of circle x2 + y2 = a2 is, , (c), , If x2 + y2 = a2 is the equation of the circle then at any point 't' of this circle (a cos t, a sint), the equation, of normal is xsint – ycost = 0., , Illustration 11 : Find the equation of the normal to the circle x2 + y2 – 5x + 2y – 48 = 0 at the point (5, 6)., Solution :, , Since normal to the circle always passes through the centre so equation of the normal will be the, , 5, , line passing through (5, 6) &  ,  1, 2, 7 , 5,  x    5y  5  14 x  35, 5/2, 2, , i.e., , y + 1 =, , , , 14x – 5y – 40 = 0, , Ans., , 52, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , (a), , E

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JEEMAIN.GURU, , JEE-Mathematics, , Illustration 12 : If the straight line ax + by = 2; a, b  0 touches the circle x2 + y2 – 2x = 3 and is normal to the, circle x2 + y2 – 4y = 6, then the values of a and b are respectively, (A) 1, –1, 2, , Solution :, , 4, (C)  , 1, 3, , (B) 1, 2, , (D) 2, 1, , 2, , Given x + y – 2x = 3, , centre is (1, 0) and radius is 2, Given x2 + y2 – 4y = 6, centre is (0, 2) and radius is, , , , | a(1)  b(0)  2|, , 2, , 10 . Since line ax + by = 2 touches the first circle, , or |(a – 2)| = [2 a 2  b 2 ], , ......... (i), a b, Also the given line is normal to the second circle. Hence it will pass through the centre of the, , , 2, , 2, , second circle., a(0) + b(2) = 2, , , , or, , 2b = 2, , or, , b=1, , Putting this value in equation (i) we get |a – 2| = 2 a 2  12, or, , a2 + 4 – 4a = 4a2 + 4, , or 3a2 + 4a = 0 or, , or, , a (3a + 4) = 0, , (a – 2)2 = 4(a2 + 1), or, , a = 0, , , 4, (a  0), 3, ,  4 ,  values of a and b are   , 1 ., Ans. (C),  3 , Illustration 13 : Find the equation of a circle having the lines x2 + 2xy + 3x + 6y = 0 as its normal and having size, just sufficient to contain the circle x(x – 4) + y(y – 3) = 0., , Solution :, , Pair of normals are (x + 2y)(x + 3) = 0, , Normals are x + 2y = 0, x + 3 = 0., Point of intersection of normals is the centre of required circle i.e. C1(–3, 3/2) and centre of given, circle is C2(2, 3/2) and radius r2 =, , 4, , 9 5, , 4 2, , Let r1 be the radius of required circle, 2, , , , r1 = C 1 C 2 + r 2 =, , 5 15, 3 3, ( 3  2) 2      , 2 2, 2, 2, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , Hence equation of required circle is x2 + y2 + 6x – 3y – 45 = 0, , E, , Do yourself - 4 :, (i), 7., , Find the equation of the normal to the circle x2 + y2 = 2x, which is parallel to the line x + 2y = 3., , CHORD OF CONTACT (T = 0) :, A line joining the two points of contacts of two tangents drawn from a point out, side the circle, is called chord of contact of that point., If two tangents PT1 & PT2 are drawn from the point P (x1, y1) to the circle, S  x2 + y2 + 2gx + 2fy + c = 0 , then the equation of the chord of contact, T1T2 is :, xx1 + yy1 + g (x + x1) + f (y + y1) + c = 0 (i.e. T = 0 same as equation of tangent)., Remember, , T1, L, , R, C, , P(x1,y1), T2, , :, , 2L R, (a), (b), , Length of chord of contact T1 T2 =, , R 2  L2, , ., , R L3, Area of the triangle formed by the pair of the tangents & its chord of contact = 2, , where, R  L2, R is the radius of the circle & L is the length of the tangent from (x1, y1) on S = 0., , 53

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JEEMAIN.GURU, , JEE-Mathematics, (c), (d), (e), ,  2R L , Angle between the pair of tangents from P(x1, y1) = tan 1  2,  L  R 2 , Equation of the circle circumscribing the triangle PT1 T2 or quadrilateral CT1PT2 is :, (x - x1) (x + g) + (y – y1) (y + f) = 0., The joint equation of a pair of tangents drawn from the point A (x1 , y1) to the circle, x2 + y2 + 2gx + 2fy + c = 0 is : SS1 = T²., Where S  x2 + y2 + 2gx + 2fy + c ; S1  x1² + y1² + 2gx1 + 2fy1 + c, T  xx1 + yy1 + g(x + x1) + f(y + y1) + c., , Illustration 14 :, Solution :, , The chord of contact of tangents drawn from a point on the circle x 2 + y 2 = a 2 to the circle, x2 + y2 = b2 touches the circle x2 + y2 = c2. Show that a, b, c are in GP., Let P(acos, asin) be a point on the circle x2 + y2 = a2., Then equation of chord of contact of tangents drawn from, P(acos, asin) to the circle x2 + y2 = b2 is axcos + aysin = b2, ..... (i), This touches the circle x2 + y2 = c2, ..... (ii), , Length of perpendicular from (0, 0) to (i) = radius of (ii), , T, , P, , R, , |0  0  b |, 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , x +y =b, , 2, , , , 2, , x +y =c, , x +y =a, , c, , (a cos   a sin ), 2, , or, , b = ac , , a, b, c are in GP., , Do yourself - 5 :, (i), , Find the equation of the chord of contact of the point (1, 2) with respect to the circle, x2 + y2 + 2x + 3y + 1 = 0, , (ii), , Tangents are drawn from the point P(4, 6) to the circle x2 + y2 = 25. Find the area of the triangle, formed by them and their chord of contact., , 8., , EQUATION OF THE CHORD WITH A GIVEN MIDDLE POINT (T = S 1 ) :, The equation of the chord of the circle S  x2 + y2 + 2gx + 2fy + c = 0 in terms of its mid point M (x1 , y1), , x1  g, (x  x1). This on simplification can be put in the form, y1  f, xx1 + yy1 + g (x + x1) + f (y + y1) + c = x12 + y12 + 2gx1 + 2fy1 + c which is designated by T = S1., is y  y1 = , , middle point is M., , Illustration 15 : Find the locus of middle points of chords of the circle x2 + y2 = a2, which subtend right angle at the, point (c, 0)., , Solution :, , Let N(h, k) be the middle point of any chord AB,, , y, , which subtend a right angle at P(c, 0)., , A, , Since APB = 90°, , , (h,k), N, , NA = NB = NP, , (since distance of the vertices from middle point of, , x, , x', P(c, 0), , the hypotenuse are equal), or, , (NA)2 = (NB)2 = (h – c)2 + (k – 0)2 ..... (i), , O, B, , But also BNO = 90°, y', , , , (OB)2 = (ON)2 + (NB)2, , , , –(NB)2 = (ON)2 – (OB)2, , or, , 2(h2 + k2) – 2ch + c2 – a2 = 0, , , , Locus of N(h, k) is 2(x2 + y2) – 2cx + c2 – a2 = 0, , –[(h – c)2 + (k – 0)2] = (h2 + k2) – a2, , , , 54, , Ans., , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , Note that : The shortest chord of a circle passing through a point ‘M’ inside the circle, is one chord whose, , E

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JEEMAIN.GURU, , JEE-Mathematics, , Illustration 16 : Let a circle be given by, , Solution :, , 2x(x – a) + y(2y – b) = 0, (a  0, b  0), Find the condition on a and b if two chords, each bisected by the x-axis, can be drawn to the circle, from (a, b/2)., The given circle is 2x(x – a) + y(2y – b) = 0, or, x2 + y2 – ax – by/2 = 0, Let AB be the chord which is bisected by x-axis at a point M. Let its co-ordinates be M(h, 0)., and, S  x2 + y2 – ax – by/2 = 0, , Equation of chord AB is T = S1, hx + 0 –, , a, b, (x  h)  (y  0)  h 2  0  ah  0, 2, 4, , a, b2, 3ah a 2 b 2, , , 0, (a + h) –, = h2 – ah  h2 –, 2, 8, 2, 2, 8, Now there are two chords bisected by the x-axis, so there must be two distinct real roots of h., , B2 – 4AC > 0, Since its passes through (a, b/2) we have ah –, , 2, , , ,  a 2 b2 ,  3a ,   0, ,   4.1. , 2, 2, 8, ,  a2 > 2b2., , Ans., , Do yourself - 6 :, , 9., , (i), , Find the equation of the chord of x2 + y2 – 6x + 10 – a = 0 which is bisected at (–2, 4)., , (ii), , Find the locus of mid point of chord of x2 + y2 + 2gx + 2ƒy + c = 0 that pass through the origin., , DIRECTOR CIRCLE :, The locus of point of intersection of two perpendicular tangents to a circle is called director circle. Let P(h,k) is, the point of intersection of two tangents drawn on the circle x2 + y2 = a2. Then the equation of the pair of, tangents is SS1= T2, i.e. (x2 + y2 – a2) (h2 + k2 – a2) = (hx + ky – a2)2, As lines are perpendicular to each other then, coefficient of x2 + coefficient of y2 = 0, , , [(h2 +k2 – a2)–h2] + [(h2 + k2 – a2)– k2] = 0, , , , , h 2 + k 2 = 2a 2, locus of (h,k) is x2 + y2 = 2a2 which is the equation of the director circle., , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , , director circle is a concentric circle whose radius is 2 times the radius of the circle., Note : The director circle of x2 + y2 + 2gx + 2fy + c = 0 is x2 + y2 + 2gx + 2fy + 2c– g2 – f2 = 0, , E, , Illustration 17 : Let P be any moving point on the circle x2 + y2 – 2x = 1, from this point chord of contact is drawn, w.r.t. the circle x2 + y2 – 2x = 0. Find the locus of the circumcentre of the triangle CAB, C being, , Solution :, , centre of the circle and A, B are the points of contact., The two circles are, (x – 1) 2 + y 2 = 1, ......... (i), (x – 1) 2 + y 2 = 2, ......... (ii), So the second circle is the director circle of the first. So APB = /2, Also ACB = /2, Now circumcentre of the right angled triangle CAB would lie on the mid point of AB, So let the point be M  (h, k), P, , 1, Now, CM = CBsin45° =, , 2, , So,, ,  1 , (h – 1) 2 + k 2 = ,  2 , , So,, , locus of M is (x – 1) 2 + y 2 =, , B, , A, M, , 2, , C, , 1, ., 2, , 55

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JEEMAIN.GURU, , JEE-Mathematics, Do yourself - 7 :, , 10., , (i), , Find the equation of the director circle of the circle (x – h)2 + (y – k)2 = a2., , (ii), , If the angle between the tangents drawn to x2 + y2 + 4x + 8y + c = 0 from (0, 0) is, , (i i i ), , of 'c', If two tangents are drawn from a point on the circle x2 + y2 = 50 to the circle x2 + y2 = 25, then find, the angle between the tangents., , , , then find value, 2, , POLE AND POLAR :, , R (h,k), , Let any straight line through the given point A(x1,y1) intersect the given circle S =0 in, two points P and Q and if the tangent of the circle at P and Q meet at the point R, then locus of point R is called polar of the point A and point A is called the pole, with, , (x1,y1), , Q, , A P, , respect to the given circle., (a), , The equation of the polar of point (x1,y1) w.r.t. circle x2 + y2 = a2 (T = 0)., Let PQR is a chord which passes through the point P(x1,y1) which intersects the, circle at points Q and R and the tangents are drawn at points Q and R meet, at point S(h,k) then equation of QR the chord of contact is x 1h + y 1k= a 2, , locus of point S(h,k) is xx1 + yy1 = a2 which is the equation of the polar., , S (h,k), R, , (x1,y1), P Q, , Note :, (i), , The equation of the polar is the T=0, so the polar of point (x1,y1) w.r.t circle, , (ii), , If point is outside the circle then equation of polar and chord of contact is same. So the chord of, contact is polar., , (iii), , If point is inside the circle then chord of contact does not exist but polar exists., , (iv), , If point lies on the circle then polar , chord of contact and tangent on that point are same., , (v), , If the polar of P w.r.t. a circle passes through the point Q, then the polar of point Q will pass through, P and hence P & Q are conjugate points of each other w.r.t. the given circle., , (vi), (vii), , If pole of a line w.r.t. a circle lies on second line. Then pole of second line lies on first line and hence, both lines are conjugate lines of each other w.r.t. the given circle., If O be the centre of a circle and P be any point, then OP is perpendicular to the polar of P., , (viii), , If O be the centre of a circle and P any point, then if OP (produce, if necessary) meet the polar of, P in Q, then OP. OQ = (radius)2, , (b), , Pole of a given line with respect to a circle, To find the pole of a line we assume the coordinates of the pole then from these coordinates we find the, polar. This polar and given line represent the same line. Then by comparing the coefficients of similar, terms we can get the coordinates of the pole. The pole of x + my + n = 0, ,  a 2 ma 2 , w.r.t. circle x + y = a will be  n , n , , , 2, , 11., , 2, , 2, , FA MILY OF CIRCLES :, (a), , The equation of the family of circles passing through the, , S1, , S2, , points of intersection of two circles, S1 = 0, (b), , &, , S2 = 0 is : S1 + K S2 = 0, , (K  –1)., S, , The equation of the family of circles passing through the point of, intersection of a circle S = 0 & a line L = 0 is given by S + KL = 0., , 56, , L, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , x2 + y2 + 2gx + 2fy + c = 0 is xx1+ yy1+ g(x + x1) + f(y + y 1)+c = 0, , E

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JEEMAIN.GURU, , JEE-Mathematics, , Do yourself - 9 :, (i), , Two circles with radius 5 touches at the point (1, 2). If the equation of common tangent is 4x + 3y = 10, and one of the circle is x2 + y2 + 6x + 2y – 15 = 0. Find the equation of other circle., , (ii), 13., , Find the number of common tangents to the circles x2 + y2 = 1 and x2 + y2 – 2x – 6y + 6 = 0., , THE ANGLE OF INTERSECTION OF TWO CIRCLES :, Definition : The angle between the tangents of two circles at the point of intersection of the two circles is called, angle of intersection of two circles. If two circles are S1  x2 + y2 + 2g1x + 2f1y + c1 = 0, S2  x2 + y2 + 2g2x + 2f2y + c2 = 0 and  is the acute angle between them, then cos  , , 2g1 g 2  2f1 f2  c 1  c 2, 2 g 12  f12  c1, , g 22  f22  c 2, , P, r1, ,  r 2  r22  d 2 , or cos    1, , 2r1 r2, , , , C1, , , d, , r2, C2, , Here r 1 and r 2 are the radii of the circles and d is the distance between their, centres., If the angle of intersection of the two circles is a right angle then such circles are, called "Orthogonal circles" and conditions for the circles to be orthogonal is 2g 1 g 2 + 2f 1 f 2 = c 1 + c 2, 14., , P(h,k), , RADICAL AXIS OF THE TWO CIRCLES (S 1 – S 2 = 0) :, (a), , Definition : The locus of a point, which moves in such a way, that the length of tangents drawn from it to the circles are equal, and is called the radical axis. If two circles are -, , A, , B, , S1  x2 + y2 + 2g1x + 2f1 y + c1 =0, S2  x2 + y2 + 2g2x + 2f2y + c2 = 0, , Radical axis, , Let P(h,k) is a point and PA,PB are length of two tangents on the circles from point P, Then from, definition -, , h 2  k 2  2g1 h  2f1 k  c 1  h 2  k 2  2g 2 h  2f2 k  c 2, locus of (h,k), 2x(g1–g 2) + 2y(f1–f 2)k + c1 – c2 = 0, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , , , E, , or 2(g1–g2) h + 2(f1–f2) k + c1 – c2 = 0, , S 1– S 2= 0, which is the equation of radical axis., Note, (i), (ii), (iii), (iv), (v), (vi), (vii), (viii), (ix), , :, To get the equation of the radical axis first of all make the coefficient of x2 and y2 =1, If circles touch each other then radical axis is the common tangent to both the circles., When the two circles intersect on real points then common chord is the radical axis of the two, circles., The radical axis of the two circles is perpendicular to the line joining the centre of two circles but, not always pass through mid point of it., Radical axis (if exist) bisects common tangent to two circles., The radical axes of three circles (taking two at a time) meet at a point., If circles are concentric then the radical axis does not always exist but if circles are not concentric, then radical axis always exists., If two circles are orthogonal to the third circle then radical axis of both circle passes through the, centre of the third circle., A system of circle, every pair of which have the same radical axis, is called a coaxial system of, circles., , 59

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JEEMAIN.GURU, , JEE-Mathematics, (b), , Radical centre :, The radical centre of three circles is the point from which length of tangents on three circles are equal i.e., the point of intersection of radical axis of the circles is the radical centre of the circles., To get the radical axis of three circles S1 =0, S2=0, S3=0 we have to solve any two, S 1 –S 2 =0, S 2 –S 3 =0, S 3 –S 1 =0, Note :, (i), , I, , The circle with centre as radical centre and radius equal to the, , T1, , C1, , =, , length of tangent from radical centre to any of the circle, will cut, (ii), , If three circles are drawn on three sides of a triangle taking them, as diameter then its orthocenter will be its radical centre., , (iii), , III, , Locus of the centre of a variable circle orthogonal to two fixed, , C2, , =, , T2, , =, , the three circles orthogonally., , T3, , II, , C3, , circles is the radical axis between the two fixed circles., (iv), , If two circles are orthogonal, then the polar of a point 'P' on first circle w.r.t. the second circle, passes through the point Q which is the other end of the diameter through P. Hence locus of a, point which moves such that its polars w.r.t. the circles S1 = 0 , S2 = 0 & S3 = 0 are concurrent, is a circle which is orthogonal to all the three circles., , Illustration 20 : A and B are two fixed points and P moves such that PA = nPB where n  1. Show that locus of P, Solution :, , is a circle and for different values of n all the circles have a common radical axis., Let A  (a, 0), B  (–a, 0) and P(h, k), so PA = nPB, , , (h – a) 2 + k 2 = n 2[(h + a) 2 + k 2 ], , , , (1 – n 2 )h 2 + (1 – n 2)k 2 – 2ah(1 + n 2 ) + (1 – n 2)a 2 = 0, , , ,  1  n2 , 2, h 2 + k 2 – 2ah ,  a 0, 1  n2 , , Hence locus of P is, ,  1  n2 , 2, x2 + y 2 – 2ax ,   a  0 , which is a circle of different values of n., 1  n2 , values of n the circles have a common radical axis., , Illustration 21 : Find the equation of the circle through the points of intersection of the circles, , Solution :, , x2 + y2 – 4x – 6y – 12 = 0 and x2 + y2 + 6x + 4y – 12 = 0 and cutting the circle, x2 + y2 – 2x – 4 = 0 orthogonally., The equation of the circle through the intersection of the given circles is, x 2 + y 2 – 4x – 6y – 12 + (–10x – 10y) = 0, , .......... (i), , where (–10x – 10y = 0) is the equation of radical axis for the circle, x 2 + y 2 – 4x – 6y – 12 = 0 and x 2 + y 2 + 6x + 4y – 12 = 0., Equation (i) can be re-arranged as, x 2 + y 2 – x(10 + 4) – y(10 + 6) – 12 = 0, It cuts the circle x 2 + y 2 – 2x – 4 = 0 orthogonally., Hence 2gg 1 + 2ff 1 = c + c 1, , , 2(5 + 2)(1) + 2(5 + 3)(0) = – 12 – 4    = – 2, , Hence the required circle is, x 2 + y 2 – 4x – 6y – 12 – 2(–10x – 10y) = 0, i.e., x 2 + y 2 + 16x + 14y – 12 = 0, , 60, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , Let n1 and n2 are two different values of n so their radical axis is x = 0 i.e. y-axis. Hence for different, , E

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JEEMAIN.GURU, , JEE-Mathematics, Do yourself - 11 :, (i), , When the circles x2 + y2 + 4x + 6y + 3 = 0 and 2(x2 + y 2) + 6x + 4y + c = 0 intersect orthogonally,, then find the value of c is, , (ii), , Write the condition so that circles x2 + y2 + 2ax + c = 0 and x2 + y2 + 2by + c = 0 touch externally., , Miscellaneous, , Illustrations, , :, , Illustration 23 : Find the equation of a circle which passes through the point (2, 0) and whose centre is the limit of, the point of intersection of the lines 3x + 5y = 1 and (2 + c)x + 5c2y = 1 as c  1., , Solution :, , Solving the equations (2 + c)x + 5c2y = 1 and 3x + 5y = 1, then, ,  1  3x , (2 + c)x + 5c2 ,  = 1, 5 , , , , x =, , , , x = lim, , , , 1  3x, y, , 5, , 1  c2, 2  c  3c 2, c 1, , 1c, 3c  2, , or, , (2 + c)x + c2 (1 – 3x) = 1, , or, , x, , (1  c)(1  c), 1c, , (3c  2)(1  c) 3c  2, , or, , x =, , 2, 5, , 6, 5  1, 5, 25, , 1, ,  2 1 , Therefore the centre of the required circle is  ,, but circle passes through (2, 0),  5 25 , 2, , , , Radius of the required circle =, , 2, , 2, ,  1, ,  0 =,   2    , , 5, 25, 2, , 64, 1, 1601, , , 25 625, 625, , 2, , 2, 1 , 1601, , , Hence the required equation of the circle is  x     y ,  , , , , 5, 25, 625, or, 25x2 + 25y2 – 20x + 2y – 60 = 0, , Ans., , Illustration 24 : Two straight lines rotate about two fixed points. If they start from their position of coincidence such, that one rotates at the rate double that of the other. Prove that the locus of their point of intersection, is a circle., Let A  (–a, 0) and B  (a, 0) be two fixed points., Let one line which rotates about B an angle  with the x-axis at any time t and at that time the, second line which rotates about A make an angle 2 with x-axis., Now equation of line through B and A are respectively, and, , y – 0 = tan(x – a), , ...... (i), , y – 0 = tan2(x + a), , ...... (ii), , From (ii),, , y, , 2 tan , (x  a ), 1  tan 2 , , , 2y, ,  x  a, , 2, 1  y, ,  x  a 2, , or, , y, , 2, A(–a, 0), , , , ,  x  a, , , , (from (i)), , 2y  x  a   x  a , , ,  x  a 2  y 2, , (x – a)2 – y2 = 2(x2 – a2), , x2 + y2 + 2ax – 3a2 = 0 which is the required locus., , 62, , O(0, 0), , , B(a, 0), , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , Solution :, , E

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JEEMAIN.GURU, , JEE-Mathematics, , Illustration 25 : If the circle x 2 + y 2 + 6x – 2y + k = 0 bisects the circumference of the circle, x2 + y2 + 2x – 6y – 15 = 0, then k =, (A) 21, (B) –21, , Solution :, , (C) 23, , (D) –23, , 2g2 (g1 – g2) + 2f2 (f1 – f2) = c1 – c2, 2(1) (3 – 1) + 2 (–3) (–1 + 3) = k + 15, , 4 – 12 = k + 15 or –8 = k + 15,  k = –23, Ans. (D), Illustration 26 : Find the equation of the circle of minimum radius which contains the three circles., S1  x2 + y2 – 4y – 5 = 0, S2  x2 + y2 + 12x + 4y + 31 = 0, S3  x2 + y2 + 6x + 12y + 36 = 0, Solution :, For S1, centre = (0, 2) and radius = 3, (0,2), For S2, centre = (–6, –2) and radius = 3, For S3, centre = (–3, –6) and radius = 3, P(a,b), let P(a, b) be the centre of the circle passing through the centres, of the three given circles, then, (a – 0)2 + (b – 2)2 = (a + 6)2 + (b + 2)2, , (a + 6)2 – a2 = (b – 2)2 – (b + 2)2, (2a + 6)6 = 2b(–4), 2  6(a  3), 3,   (a  3), 8, 2, again (a – 0)2 + (b – 2)2 = (a + 3)2 + (b + 6)2, , b =, , , , (a + 3)2 – a2 = (b – 2)2 – (b + 6)2, (2a + 3)3 = (2b + 4) (– 8),  3, , (2a + 3)3 = –16   (a  3)  2 ,  2, , , 6a + 9 = –8(–3a – 5), 6a + 9 = 24a + 40, 18a = –31, a = , , 31, 23, ,b, 18, 12, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , 2, , E, , 2, ,  31   23, , radius of the required circle = 3   ,     12  2 , 18, ,  , , 2, , = 3, 2, , 5, 949, 36, 2, , 31  , 23 , 5, , , , y ,  3 , 949 , equation of the required circle is  x , , , 18  , 12 , 36, , , , 2, 2, Illustration 27 : Find the equation of the image of the circle x + y + 16x – 24y + 183 = 0 by the line mirror, 4x + 7y + 13 = 0., Solution :, Centre of given circle = (–8, 12), radius = 5, the given line is 4x + 7y + 13 = 0, let the centre of required circle is (h, k), since radius will not change. so radius of required circle is 5., Now (h, k) is the reflection of centre (–8, 12) in the line 4x + 7y + 13 = 0, , ,  8  h 12  k , ,, Co-ordinates of A = , 2 ,  2, , 4( 8  h) 7(12  k ), ,  13  0, , 2, 2, –32 + 4h + 84 + 7k + 26 = 0, 4h + 7k + 78 = 0, .........(i), , 63, , (–8,12), , 0, 3=, y+1, 7, +, 4x, , A, (h,k)

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JEEMAIN.GURU, , JEE-Mathematics, k  12 7, , h 8, 4, 4k – 48 = 7h + 56, 4k = 7h + 104, .........(ii), solving (i) & (ii), h = –16, k = –2, , required circle is (x + 16)2 + (y + 2)2 = 52, Illustration 28 : The circle x2 + y2 – 6x – 10y + k = 0 does not touch or intersect the coordinate axes and the point, (1, 4) is inside the circle. Find the range of the value of k., Solution :, Since (1, 4) lies inside the circle, , S1 < 0, , (1)2 + (4)2 – 6(1) – 10(4) + k < 0, , k < 29, Also centre of given circle is (3, 5) and circle does not touch or intersect the coordinate axes, , r < CA & r < CB, r C(3,5), CA = 5, B, r, CB = 3, , r < 5 &, r < 3, , r < 3 or r2 < 9, r2 = 9 + 25 – k, A, r2 = 34 – k, , 34 – k < 9, k > 25, , k  (25, 29), , Also, , Illustration 29 : The circle x2 + y2 – 4x – 8y + 16 = 0 rolls up the tangent to it at (2 +, Solution :, , 3 , 3) by 2 units, find the, , equation of the circle in the new position., Given circle is x2 + y2 – 4x – 8y + 16 = 0, let P  (2 +, , B, , 3 , 3), , Equation of tangent to the circle at P(2 + 3 , 3) will be, 2, , 3 )x + 3y – 2(x + 2 + 3 ) – 4(y + 3) + 16 = 0, , or, slope =, , 3x–y –2 3 =0, A, (2,4), , , tan = 3,  = 60°, line AB is parallel to the tangent at P, , coordinates of point B = (2 + 2cos60°, 4 + 2sin60°), 3, , P(2+ 3,3), , thus B = (3, 4 + 3 ), radius of circle = 2 2  4 2  16  2, , equation of required circle is (x – 3)2 + (y – 4 – 3 )2 = 22, Illustration 30 : A fixed circle is cut by a family of circles all of which, pass through two given points A(x1, y1) and, B(x2, y2). Prove that the chord of intersection of the fixed circle with any circle of the family passes, through a fixed point., , Solution :, , Let S = 0 be the equation of fixed circle, , S1=0, S=0, , let S1 = 0 be the equation of any circle through A and B, , A(x1,y1), , which intersect S = 0 in two points., L  S – S1 = 0 is the equation of the chord of intersection, of S = 0 and S1 = 0, let L1 = 0 be the equation of line AB, , 64, , B, (x2,y2), L=0, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , (2 +, , E

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JEEMAIN.GURU, , JEE-Mathematics, , let S2 be the equation of the circle whose diametrical ends are A(x1, y1) & B(x2, y2), then S1  S2 – L1 = 0, , , L  S – (S2 – L1) = 0, , or, , L  L' + L1 = 0, , or, , L  (S – S2) + L1 = 0, ........(i), , (i) implies each chord of intersection passes through the fixed point, which is the point of intersection, of lines L' = 0 & L1 = 0. Hence proved., , Illustration 31 : Let L1 be a straight line through the origin and L2 be the straight line x + y = 1. If the intercepts, made by the circle x2 +, equations can represent, (A) x + y = 0, , Solution :, , y2 – x + 3y = 0 on L1 & L2 are equal , then which of the following, L1?, , (B) x – y = 0, , (C) x + 7y = 0, , (D) x – 7y = 0, , Let L1 be y = mx, 3, 1, lines L1 & L2 will be at equal distances from centre of the circle centre of the circle is  ,  , 2, 2, , , 1, 3, m, 2, 2, , , , , , 1  m2, , 1 3,  1, 2 2, , (m  3)2, , , , (1  m 2 ), , 2, , , , 7m2 – 6m – 1 = 0, , , , m = 1, m = –, , 1, 7, , 8, , , , (m – 1) (7m + 1) = 0, , , , y = x, 7y + x = 0, , Ans. (B, C), , ANSWERS FOR DO YOURSELF, 1 :, , (i), , 5, 3, 3 10, Centre  ,   , Radius, 4, 4, 4, , (ii), , 17(x2 + y2) + 2x – 44y = 0, , p, p, ( 1  2 cos ) ; y  ( 1  2 sin ), (iv) x2 + y2 + 6x – 2y – 51 = 0, 2, 2, (1, 2) lie inside the circle and the point (6, 0) lies outside the circle, min = 0, max = 6, power = 0, xcos + ysin = a(1 + cos), ( i i ) 4x – 3y + 7 = 0 & 4x – 3y – 43 = 0, , (i i i ) x =, 2 :, , (i), , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , (ii), , E, , 3 :, , (i), , (i i i ) 5x + 12y = ±26 ;, 4 :, , (i), , 5 :, , (i), , 6 :, , (i), , 24 ,  10,   13 ,  13 , , , , (iv) 1, , x + 2y = 1, 4x + 7y + 10 = 0, 5x – 4y + 26 = 0, 2, , 2, , 2, , (ii), , 405 3, sq. units, 52, , (ii), , x2 + y2 + gx + ƒy = 0, , (ii), , 10, , 7 :, , (i), , (x – h) + (y – k) = 2a, , (i i i ) angle between the tangents = 90°, , 8 :, , (ii), , x2  y2 , , 9 :, , (i), , (x – 5)2 + (y – 5)2 = 25, , (ii), , 4, , 10 : ( i ), , 135°, , (ii), , x + 2y = 2, , 11 : ( i ), , 18, , (ii), , a–2 + b –2 = c –1, , 10x 10y 12, , ,  0 (i i i ) x2 + y2 + 4x – 7y + 5 = 0, 7, 7, 7, , 65, , (i i i ) (1, 2)

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JEEMAIN.GURU, 15., , Line 3x + 4y = 25 touches the circle x2 + y2 = 25 at the point (A) (4, 3), , 16., , 17., , (B) (3, 4), , (C) (–3, –4), , (D) none of these, , The equations of the tangents drawn from the point (0,1) to the circle x2 + y2 – 2x + 4y = 0 are (A) 2x – y + 1 = 0, x + 2y – 2 = 0, , (B) 2x – y – 1 = 0, x + 2y – 2 = 0, , (C) 2x – y + 1 = 0, x + 2y + 2 = 0, , (D) 2x – y – 1 = 0, x + 2y + 2 = 0, , The greatest distance of the point P(10,7) from the circle x2 + y2 – 4x – 2y – 20 = 0 is (A) 5, , 18., , JEE-Mathematics, , (B) 15, , (C) 10, , (D) None of these, ,  3 3 , ,, The equation of the normal to the circle x2 + y2 = 9 at the point ,  is  2 2, , 2, (B) x + y = 0, (C) x – y = 0, (D) none of these, 3, The parametric coordinates of any point on the circle x2 + y2 – 4x – 4y = 0 are(A) (–2 + 2cos, –2 + 2 sin), (B) (2 + 2cos, 2 + 2 sin), (A) x – y , , 19., , (C) (2 + 2 2 cos, 2 + 2 2 sin), 20., , The length of the tangent drawn from the point (2,3) to the circles 2(x2 + y2) – 7x + 9y – 11 = 0 (A) 18, , 21., , (D) (–2 + 2 2 cos, –2 + 2 2 sin), , (B) 14, , (C), , (D), , 14, , 28, , A pair of tangents are drawn from the origin to the circle x2 + y2 + 20(x + y) + 20 = 0. The equation of the pair, of tangents is (A) x2 + y2 + 5xy = 0, , 22., , (C) 2x2 + 2y2 + 5xy = 0, , (D) 2x2 + 2y2 – 5xy = 0, , Tangents are drawn from (4, 4) to the circle x 2 + y 2 – 2x – 2y – 7 = 0 to meet the circle at A and B., The length of the chord AB is (A) 2 3, , 23., , (B) x2 + y2 + 10xy = 0, , (B) 3 2, , (C) 2 6, (D) 6 2, 2, 2, The angle between the two tangents from the origin to the circle (x –7) + (y + 1) = 25 equals -, , , , , (B), (C), (D) none, 2, 3, 4, Pair of tangents are drawn from every point on the line 3x + 4y = 12 on the circle x2 + y2 = 4. Their variable, chord of contact always passes through a fixed point whose co-ordinates are (A), , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , 24., , E, ,  4 3, (A)  , , 3 4, 25., , 26., , 28., ,  4, (D)  1, , 3, , (C) (1, 1), , The locus of the mid-points of the chords of the circle x 2 + y 2 – 2x – 4y – 11 = 0 which subtend 60° at, the centre is (A) x 2 + y 2 – 4x – 2y – 7 = 0, , (B) x 2 + y 2 + 4x + 2y – 7 = 0, , (C) x 2 + y 2 – 2x – 4y – 7 = 0, , (D) x 2 + y 2 + 2x + 4y + 7 = 0, , The locus of the centres of the circles such that the point (2,3) is the mid point of the chord 5x + 2y = 16 is (A) 2x – 5y + 11 = 0, , 27., ,  3 3, (B)  , , 4 4, , (B) 2x + 5y – 11 = 0, , (C) 2x + 5y + 11 = 0, 2, , (D) none, 2, , The locus of the centre of a circle which touches externally the circle, x + y – 6x – 6y + 14 = 0 and, also touches the y-axis is given by the equation (A) x 2 – 6x – 10y + 14 = 0, , (B) x 2 – 10x – 6y + 14 = 0, , (C) y 2 – 6x – 10y + 14 = 0, , (D) y 2 – 10x – 6y + 14 = 0, , The equation of the circle having the lines y2 – 2y + 4x – 2xy = 0 as its normals & passing through the point, (2,1) is (A) x2 + y2 – 2x – 4y + 3 = 0, , (B) x2 + y2 – 2x + 4y – 5 = 0, , (C) x2 + y2 + 2x + 4y –13 = 0, , (D) none, , 67

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JEEMAIN.GURU, , JEE-Mathematics, , 29., , A circle is drawn touching the x-axis and centre at the point which is the reflection of (a, b) in the line, y – x = 0. The equation of the circle is -, , 30., , (A) x 2 + y 2 – 2bx – 2ay + a 2 = 0, , (B) x 2 + y 2 – 2bx – 2ay + b 2 = 0, , (C) x 2 + y 2 – 2ax – 2by + b 2 = 0, , (D) x 2 + y 2 – 2ax – 2by + a 2 = 0, , The length of the common chord of circles x2 + y2 – 6x – 16 = 0 and x2 + y2 – 8y – 9 = 0 is (A) 10 3, , 31., , (B) 5 3, , (B) 3, , If the circle, , x2, , +, , y2, , = 9 touches the circle, , (A) –27, 33., , (D) none of these, , The number of common tangents of the circles x2 + y2 – 2x – 1 = 0 and x2 + y2 – 2y – 7 = 0 (A) 1, , 32., , (C) 5 3 / 2, , (C) 2, x2, , +, , y2, , + 6y + c = 0, then c is equal to -, , (B) 36, 2, , (D) 4, , (C) –36, , 2, , 2, , (D) 27, , 2, , If the two circles, x + y + 2g 1 x + 2f 1 y = 0 and x + y + 2g 2 x + 2f 2 y = 0 touches each other, then -, , f1, f2, (B) g =, (C) f 1 f 2 = g 1 g 2, (D) none, g2, 1, The tangent from the point of intersection of the lines 2x – 3y + 1 = 0 and 3x – 2y – 1 = 0 to the circle, x 2 + y 2 + 2x – 4y = 0 is (A) x + 2y = 0, x – 2y + 1 = 0, (B) 2x – y – 1 = 0, (C) y = x, y = 3x – 2, (D) 2x + y + 1 = 0, The locus of the centers of the circles which cut the circles x 2 + y 2 + 4x – 6y + 9 = 0 and, x 2 + y 2 – 5x + 4y – 2 = 0 orthogonally is (A) f 1 g 1 = f 2 g 2, , 34., , 35., , (A) 9x + 10y – 7 = 0, , (B) x – y + 2 = 0, , (C) 9x – 10y + 11 = 0, , (D) 9x + 10y + 7 = 0, , SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS), 36., , Equation, , x  x1 y  y 1, ,  r, may represents cos , sin , , (A) Equation of straight line, if  is constant and r is variable., (B) Equation of a circle, if r is constant &  is variable., (C) A straight line passing through a fixed point & having a known slope., (D) A circle with a known centre and given radius., If r represent the distance of a point from origin &  is the angle made by line joining origin to that point from, line x-axis, then r = |cos| represents (A) two circles of radii, , 38., ,  1 , &  , 0,  2 , , (C) two circles touching each other at the origin., (D) pair of straight line, If the circle C1 : x2 + y2 = 16 intersects another circle C2 of radius 5 in such a manner that the common chord, is of maximum length 8 has a slope equal to, 12 , 9, (A)  , , 5 , 5, , 39., , 1 , (B) two circles centred at  , 0 , 2 , , 1, each., 2, , 3, , then coordinates of centre of C2 are 4, ,  9 12 , (B)   ,, ,  5 5 , ,  9 12 , (C)  ,, , 5 5 , , For the equation x2 + y2 + 2x + 4 = 0 which of the following can be true (A) It represents a real circle for all   R., (B) It represents a real circle for || > 2., (C) The radical axis of any two circles of the family is the y-axis., (D) The radical axis of any two circles of the family is the x-axis., , 68, , 12 ,  9, (D)   , , 5 ,  5, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , 37., , E

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JEEMAIN.GURU, 40., , JEE-Mathematics, , If y = c is a tangent to the circle x2 + y2 – 2x + 2y – 2 = 0, then the value of c can be (A) 1, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , 41., , E, , (B) 3, , (C) –1, , (D) –3, , For the circles S1  x2 + y2 – 4x – 6y – 12 = 0 and S2  x2 + y2 + 6x + 4y – 12 = 0 and the line L  x  y  0, (A) L is common tangent of S1 and S2, (B) L is common chord of S1 and S2, (C) L is radical axis of S1 and S2, (D) L is perpendicular to the line joining the centre of S1 & S2, , ANSWER, , CHE CK Y OU R G R ASP, , KEY, , EXERCISE-1, , Que., , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , Ans., , C, , B, , B, , A, , C, , D, , A, , D, , A, , A, , Que., , 11, , 12, , 13, , 14, , 15, , 16, , 17, , 18, , 19, , 20, , Ans., , B, , C, , D, , C, , B, , A, , B, , C, , C, , C, , Que., , 21, , 22, , 23, , 24, , 25, , 26, , 27, , 28, , 29, , 30, , Ans., , C, , B, , A, , D, , C, , A, , D, , A, , B, , B, , Que., , 31, , 32, , 33, , 34, , 35, , 36, , 37, , 38, , 39, , 40, , Ans., , A, , A, , B, , B, , C, , A,B,C,D, , A, B, C, , A,B, , B,C, , A,D, , Que., , 41, , Ans., , B,C,D, , 69

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JEEMAIN.GURU, , JEE-Mathematics, , EXERCISE - 02, , BRAIN TEASERS, , SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS), 1., , 2., ,  1   1   1,  1, If  a,  ,  b,  ,  c,  &  d,  are four distinct points on a circle of radius 4 units then, abcd =,  a   b  c,  d, (A) 4, (B) 1/4, (C) 1, (D) 16, What is the length of shortest path by which one can go from (–2, 0) to (2, 0) without entering the interior of, circle, x2 + y2 = 1 ?, 2, , (C) 2 3 , (D) none of these, 3, 3, Three equal circles each of radius r touch one another. The radius of the circle touching all the three given, , (A) 23, 3., , 3, , (B), , circles internally is (A) (2  3 )r, 4., 5., , (B), , 2  3  r, , (C), , 2  3  r, , (D) (2  3 )r, 3, 3, If a2 + b2 = 1, m2 + n2 = 1, then which of the following is true for all values of m, n, a, b (A) |am + bn|  1, (B) |am – bn|  1, (C) |am + bn|  1, (D) |am – bn|  1, Circles are drawn touching the co-ordinate axis and having radius 2, then (A) centre of these circles lie on the pair of lines y2 – x2 = 0, (B) centre of these circles lie only on the line y = x, (C) Area of the quadrilateral whose vertices are centre of these circles is 16 sq.unit, (D) Area of the circle touching these four circles internally is 4 (3  2 2 ), , 6., , The distance between the chords of contact of tangents to the circle x2 + y2 + 2gx + 2fy + c = 0 from the origin and, from the point (g,f) is -, , 7., , 8., , 9., , g2  f2, , (B), , g2  f2  c, 2, , (C), , g2  f2  c, 2 g2  f 2, , (D), , g2  f2  c, 2 g2  f2, , x2 + y2 + 6x = 0 and x2 + y2 – 2x = 0 are two circles, then (A) They touch each other externally, (B) They touch each other internally, (C) Area of triangle formed by their common tangents is 33 sq. units., (D) Their common tangents do not form any triangle., Tangents are draw n to the circle x 2 + y 2 = 1 at the poi nt s where it is met by the circle s,, x2 + y2 – ( + 6)x + (8 – 2)y – 3 = 0,  being the variable. The locus of the point of intersection of these, tangents is (A) 2x – y + 10 = 0, (B) x + 2y – 10 = 0, (C) x – 2y + 10 = 0, (D) 2x + y – 10 = 0, 3 circle of radii 1, 2 and 3 and centres at A, B and C respectively, touch each other. Another circle whose centre, is P touches all these 3 circles externally and has radius r. Also PAB   & PAC   (A) cos  , , 10., , 3r, 3 (1  r ), , (B) cos  , , Slope of tangent to the circle (x –, , r)2, , +, , 2r, 2 (1  r ), y2, , =, , r2, , (C) r , , 6, 23, , (D) r , , 12., , 23, , at the point (x, y) lying on the circle is -, , x, rx, y2  x2, y2  x2, (B), (C), (D), y r, y, 2 xy, 2 xy, The circle passing through the distinct points (1,t) , (t,1) & (t,t) for all values of ‘t’, passes through the point (A) (–1, –1), (B) (–1, 1), (C) (1, –1), (D) (1,1), AB is a diameter of a circle. CD is a chord parallel to AB and 2CD = AB. The tangent at B meets the line AC, produced at E then AE is equal to (A), , 11., , 6, , (A) AB, , (B), , (C) 2 2AB, , 2 AB, , 70, , (D) 2AB, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , (A), , E

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JEEMAIN.GURU, 13., , JEE-Mathematics, , The locus of the mid points of the chords of the circle x2 + y2 – ax – by = 0 which subtend a right angle at, ,  a b,  2 , 2  is (B) ax + by = a2 + b2, , (A) ax + by = 0, , 14., , a 2  b2, a 2  b2, 2, 2, 0, 0, (D) x  y  ax  by , 8, 8, A variable circle is drawn to touch the x-axis at the origin. The locus of the pole of the straight line, x + my + n = 0 w.r.t. the variable circle has the equation (A) x(my – n) – y2 = 0, (B) x(my + n) – y2 = 0, (C) x(my – n) + y2 = 0, (D) none, , 15., , (6,0) , (0,6) and (7,7) are the vertices of a triangle. The circle inscribed in the triangle has the equation -, , 2, 2, (C) x  y  ax  by , , (A) x2 + y2 – 9x + 9y + 36 = 0, x2, , 16., 17., , (B) x2 + y2 – 9x – 9y + 36 = 0, , y2, , (C), +, + 9x – 9y + 36 = 0, (D) x2 + y2 – 9x – 9y – 36 = 0, Number of points (x, y) having integral coordinates satisfying the condition x 2 + y 2 < 25 is (A) 69, (B) 80, (C) 81, (D) 77, The centre(s) of the circle(s) passing through the points (0, 0), (1, 0) and touching the circle, x 2 + y 2 = 9 is/are -, , 3, (A)  ,, 2, 18., , 19., , 1, 1 3,  1 1/2 , 1, 1/2 , (B)  , , (C)  , 2 , (D)  ,  2 , , 2, 2 2, 2, 2, The equation(s) of the tangent at the point (0, 0) to the circle, making intercepts of length 2a and 2b units, on the co-ordinate axes, is (are) (A) ax + by = 0, (B) ax – by = 0, (C) x = y, (D) bx + ay = 0, Tangents are drawn to the circle x 2 + y 2 = 50 from a point 'P' lying on the x-axis. These tangents meet, the y-axis at points 'P1 ' and 'P 2 '. Possible co-ordinates of 'P' so that area of triangle PP 1 P 2 is minimum, is/are (A) (10, 0), , 20., 21., , (B) (10 2, 0), , (C) (–10, 0), 2, , 2, , The tangents drawn from the origin to the circle x + y – 2rx – 2hy + h = 0 are perpendicular if (A) h = r, (B) h = –r, (C) r 2 + h 2 = 1, (D) r 2 + h 2 = 2, The common chord of two intersecting circles C1 and C2 can be seen from their centres at the angles of 90° &, 60° respectively. If the distance between their centres is equal to, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , 22., , E, , 3 + 1 then the radii of C1 and C2 are -, , (A) 3 and 3, (B) 2 and 2 2, (C) 2 and 2, (D) 2 2 and 4, In a right triangle ABC, right angled at A, on the leg AC as diameter, a semicircle is described. The chord, joining A with the point of intersection D of the hypotenuse and the semicircle, then the length AC equals to (A), , 23., , (D) ( 10 2, 0), , 2, , AB . AD, 2, , AB  AD, , (B), , 2, , AB . AD, AB  AD, , (C), , (D), , AB . AD, , AB . AD, AB 2  AD 2, , A circle touches a straight line x + my + n = 0 and cuts the circle x 2 + y 2 = 9 orthogonally. The locus, of centres of such circles is (A) (x + my + n) 2 = ( 2 + m 2 ) (x 2 + y 2 – 9), (B) (x + my – n) 2 = ( 2 + m 2) (x 2 + y 2 – 9), (C) (x + my + n) 2 = ( 2 + m 2 ) (x 2 + y 2 + 9), (D) none of these, , ANSWER, , BRAIN TEASER S, , KEY, , EXERCISE-2, , Que., , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , Ans., , C, , C, , B, , A,D, , A ,C , D, , C, , A,C, , A, , A, B, C, , B,C, , Que., , 11, , 12, , 13, , 14, , 15, , 16, , 17, , 18, , 19, , 20, , Ans., , D, , D, , C, , A, , B, , A, , C,D, , A,B, , A,C, , A,B, , Que., , 21, , 22, , 23, , Ans., , C, , D, , A, , 71

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JEEMAIN.GURU, , JEE-Mathematics, , EXERCISE - 03, , MISCELLANEOUS TYPE QUESTIONS, , MATCH THE COLUMN, Following question contains statements given in two columns, which have to be matched. The statements in, Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given, statement in Column-I can have correct matching with ONE OR MORE statement(s) in Column-II., 1., , Column-I, (A), , (B), , (C), , (D), , Column-II, , If point of intersection and number of common tangents of two, circles x 2 + y 2 – 2x – 6y + 9 = 0 and x 2 + y 2 + 6x – 2y + 1 = 0, are and µ respectively, then, If point of intersection and number of tangents of two circles, x 2 + y 2 – 6x = 0 and x 2 + y 2 + 2x = 0 are  and µ, respectively, then, If the straight line y = mx  m  I touches or lies outside, the circle x 2 + y 2 – 20y + 90 = 0 and the maximum and, minimum values of |m| are µ &  respectively then, If two circle x 2 + y 2 + px + py – 7 = 0 and, x 2 + y 2 – 10x + 2py + 1 = 0 cut orthogonally and, the value of p are  & µ respectively then, , (p), , µ –  = 3, , (q), , µ +  = 5, , (r), , µ –  = 4, , (s), , µ +  = 4, , ASSERTION & REASON, These questions contains, Statement-I (assertion) and Statement-II (reason)., (A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I., (B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I., (C) Statement-I is true, Statement-II is false., (D) Statement-I is false, Statement-II is true., 1., , Consider two circles C 1  x 2 + y 2 + 2x + 2y – 6 = 0 & C 2  x 2 + y 2 + 2x + 2y – 2 = 0., Statement-I : Two tangents are drawn from a point on the circle C 1 to the circle C 2, then tangents always, perpendicular., Because, , 2., , (C) C, , (D) D, , Statement-I : The line (x – 3)cos + (y –3)sin = 1 touches a circle (x – 3) 2 + (y – 3) 2 = 1 for all values, of ., Because, Statement-II : x cos + y sin = a is a tangent of circle x 2 + y 2 = a 2 for all values of ., , 3., , (A) A, , (B) B, , Consider the circles C 1 , , x2, , +, , (C) C, y2, , – 6x – 4y + 9 = 0 and C 2 , , (D) D, x2, , +, , y2, , – 8x – 6y + 23 = 0., , Statement-I : Circle C 1 bisects the circumference of the circle C 2 ., Because, Statement-II : Centre of C 1 lie on C 2 ., (A) A, (B) B, 4., , Statement-I : Circles, , x2, , +, , y2, , = 4 and, , x2, , (C) C, +, , y2, , (D) D, , – 8x + 7 = 0 intersect each other at two distinct points, , Because, Statement-II : Circles with centres C 1 and C 2 and radii r 1 and r 2 intersect at two distinct points, if, |C 1 C 2 | < r 1 + r 2, (A) A, , (B) B, , (C) C, , 72, , (D) D, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , Statement-II : C 1 is the director circle of C 2 ., (A) A, (B) B, , E

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JEEMAIN.GURU, , JEE-Mathematics, , COMPREHENSION BASED QUESTIONS, Comprehension # 1 :, , 1., 2., 3., , 4., , 5., , Let A  (–3, 0) and B  (3, 0) be two fixed points and P moves on a plane such that PA = nPB (n  0)., On the basis of above information, answer the following questions :, If n  1, then locus of a point P is (A) a straight line, (B) a circle, (C) a parabola, (D) an ellipse, If n = 1, then the locus of a point P is (A) a straight line, (B) a circle, (C) a parabola, (D) a hyperbola, If 0 < n < 1, then (A) A lies inside the circle and B lies outside the circle, (B) A lies outside the circle and B lies inside the circle, (C) both A and B lies on the circle, (D) both A and B lies inside the circle, If n > 1, then (A) A lies inside the circle and B lies outside the circle (B) A lies outside the circle and B lies inside the circle, (C) both A and B lies on the circle, If locus of P is a circle, then the circle (A) passes through A and B, (C) passes through A but does not pass through B, , (D) both A and B lies inside the circle, (B) never passes through A and B, (D) passes through B but does not pass through A, , Comprehension # 2 :, P is a variable point of the line L = 0. Tangents are drawn to the circle x 2 + y 2 = 4 from P to touch it, at Q and R. The parallelogram PQSR is completed., On the basis of above information, answer the following questions :, 1., , If L  2x + y – 6 = 0, then the locus of circumcetre of PQR is (A) 2x – y = 4, , 2., , (6 ) 3 / 2,  sq. units, 25, , (B), , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , E, , (D) x + 2y = 3, , (24) 3 / 2, sq. units, 25, , (C), , 48 6, sq. units, 25, ,  51 68 , , , (B)  , 25 25 , ,  46 68 , , , (C)  , 25 25 , , ANSWER, , M ISCEL L AN E OU S TYP E Q U ESTION, , KEY, , EXERCISE-3, , Matc h th e C o lu mn, A s s er ti o n & R eas o n, 1. A, , , , 192 6, sq. units, 25, ,  68 51 , , , (D)  , 25 25 , , 1 . (A)(r, s) ; (B) (s); (C)  (p); (D)  (q), , , (D), , If P  (3, 4), then coordinate of S is -, ,  46 63 , , , (A)  , 25 25 , , , , (C) x – 2y = 4, , If P  (6, 8), then the area of QRS is (A), , 3., , (B) 2x + y = 3, , 2. A, , C o mp r eh e ns i o n, , 3. B, , B as ed, , 4. C, , Qu es ti o ns, , Comprehensi on # 1 :, , 1. B, , 2. A, , 3. A, , Comprehensi on # 2 :, , 1. B, , 2. D, , 3. B, , 73, , 4. B, , 5. B

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JEEMAIN.GURU, , JEE-Mathematics, , EXERCISE - 04 [A], , CONCEPTUAL SUBJECTIVE EXERCISE, , 1., , Find the equations of the circles which have the radius, , 13 & which touch the line 2x  3y + 1 = 0 at (1, 1)., , 2., , (x1, y1) & (x2, y2) are the ends of a diameter of a circle such that x1 & x2 are the roots of, , ax² + bx + c = 0 &, , y1 & y2 are roots of py² + qy + r = 0. Find the equation of the circle, its centre & radius., 3., , If the lines a1 x + b1 y + c1 = 0 & a2 x + b2 y + c2 = 0 cut the coordinate axes in concyclic points. Prove that, a1 a2 = b1 b 2., , 4., , A (–a, 0) ; B (a, 0) are fixed points. C is a point which divides internally AB in a constant ratio tan. If AC &, CB subtend equal angles at P, prove that the equation of the locus of P is x² + y² + 2ax sec2 + a² = 0., , 5., , Let A be the centre of the circle x² + y²  2x  4y  20 = 0. Suppose that the tangents at the points, B(1 , 7) & D(4 , -2) on the circle meet at the point C. Find the area of the quadrilateral ABCD., , 6., , Determine the nature of the quadrilateral formed by four lines 3x + 4y – 5 = 0; 4x – 3y – 5 = 0; 3x + 4y + 5 = 0, and 4x – 3y + 5 = 0. Find the equation of the circle inscribed and circumscribing this quadrilateral., , 7., , A variable circle passes through the point A (a, b) & touches the x-axis ; show that the locus of the other end, of the diameter through A is (x  a)² = 4by., , 8., , A circle is drawn with its centre on the line x + y = 2 to touch the line 4x – 3y + 4 = 0 and pass through the point, (0, 1). Find its equation., , 9., , Obtain the equations of the straight lines passing through the point A(2, 0) & making 45° angle with the tangent, at A to the circle (x + 2)² + (y  3)² = 25. Find the equations of the circles each of radius 3 whose centres are, on these straight lines at a distance of 5 2 from A., , 10., , Suppose the equation of the circle which touches both the coordinates axes and passes through the point with, abscissa – 2 and ordinate 1 has the equation x2 + y2 + Ax + By + C = 0, find all the possible ordered triplet (A, B, C)., , 11., , The foot of the perpendicular from the origin to a variable tangent of the circle x2 + y2  2x = 0 is N., Find the equation of the locus of N., , 12., , The line x + my + n = 0 intersects the curve ax2 + 2hxy + by2 = 1 at the point P and Q. The circle on PQ as, diameter passes through the origin. Prove that n2(a + b) = 2 + m2., , 13., , Find the equation of the circle which passes through the point (1, 1) & which touches the circle, x² + y² + 4x  6y  3 = 0 at the point (2, 3) on it., , 14., , A circle S = 0 is drawn with its centre at (–1, 1) so as to touch the circle x2 + y2 – 4x + 6y – 3 = 0 externally., , 15., , Find the equation of the circle which cuts each of the circles x² + y² = 4 , x² + y²  6x  8y + 10 = 0, & x² + y² + 2x  4y  2 = 0 at the extremities of a diameter., , 16., , If the line x sin – y + a sec = 0 touches the circle with radius 'a' and centre at the origin then find the most, general values of '' and sum of the values of '' lying in [0, 100]., , 17., , Let a circle be given by 2x(x – a) + y(2y – b) = 0, (a  0, b  0). Find the condition on a & b if two chords, each, , 18., ,  b, bisected by the x-axis, can be drawn to the circle from the point  a,  .,  2, Find the equation of a line with gradient 1 such that the two circles x2 + y2 = 4 and x2 + y2 – 10x – 14y + 65 = 0, intercept equal length on it., , 19., , Find the equations of straight lines which pass through the intersection of the lines x  2y  5 = 0, 7x + y = 50, & divide the circumference of the circle x² + y² = 100 into two arcs whose lengths are in the ratio 2 : 1., , 20., , Find the locus of the middle points of portions of the tangents to the circle x2 + y 2 = a2 terminated by the coordinate, axes., , 21., , Show that the equation of a straight line meeting the circle x2 + y2 = a2 in two points at equal distances 'd' from, a point (x1 , y1) on its circumference is xx1 + yy1  a2 +, , 74, , d2, = 0., 2, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , Find the intercept made by the circle S = 0 on the coordinates axes., , E

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JEEMAIN.GURU, 22., , JEE-Mathematics, , A point moving around circle (x + 4)2 + (y + 2)2 = 25 with centre C broke away from it either at the point A or, point B on the circle and move along a tangent to the circle passing through the point D(3, –3)., Find the following :, (a), , Equation of the tangents at A and B., , (b), , Coordinates of the points A and B., , (c), , Angle ADB and the maximum and minimum distances of the point D from the circle., , (d), , Area of quadrilateral ADBC and the DAB., , (e), , Equation of the circle circumscribing the DAB and also the intercepts made by the this circle on the, coordinates axes., , 23., , Show that the equation x2 + y2  2x  2 y  8 = 0 represents, for different values of , a system of circles, passing through two fixed points A, B on the x  axis, and find the equation of that circle of the system the, tangents to which at A & B meet on the line x + 2y + 5 = 0., , 24., , Through a fixed point (h, k) secants are drawn to the circle x 2 + y 2 = r 2 . Show that the locus of the, mid-points of the secants intercepted by the circle is x 2 + y 2 = hx + ky., , 25., , A triangle has two of its sides along the coordinate axes, its third side touches the circle x² + y²  2ax  2ay + a² = 0., Prove that the locus of the circumcentre of the triangle is : a²  2a (x + y) + 2xy = 0., , 26., , Find the equations to the four common tangents to the circles x² + y² = 25 and (x  12)² + y² = 9., , 27., , Show that the locus of the centres of a circle which cuts two given circles orthogonally is a straight line &, hence deduce the locus of the centre of the circles which cut the circles x² + y² + 4x  6y + 9 = 0, , &, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , x² + y²  5x + 4y + 2 = 0 orthogonally., , E, , CON CEP TUAL SU BJ ECTIVE E X ER CISE, , 1 . x² + y²  6x + 4y = 0, , or, , ANSWER, , KEY, , EXERCISE-4(A), ,  b, q, 2. x² + y² +   x +   y +  c  r  = 0,  a p , a,  p, , x² + y² + 2x  8y + 4 = 0, , 5. 75 sq.units 6 . square of side 2; x2 + y2 = 1; x2 + y2 = 2, 8 . x2 + y2 – 2x – 2y + 1 = 0 or x2 + y2 – 42x + 38y – 39 = 0, 9 . x  7y = 2, 7x + y = 14; (x  1)2 + (y  7)2 = 32; (x  3)2 + (y + 7)2 = 32; (x  9)2 + (y  1)2 = 32; (x + 5)2 + (y + 1)2 = 32, 10. x2 + y2 + 10x – 10y + 25 = 0 or x2 + y2 + 2x – 2y + 1 = 0, (10, –10, 25) (2, –2, 1), 11. (x² + y2  x)2 = x2 + y2, 1 3 . x² + y² + x  6y + 3 = 0, 16.  = n, 5050, 17. a² > 2b 2, 18. 2x – 2y – 3 = 0, 20. a2(x 2 + y 2) = 4x 2y 2, , 1 4 . zero, zero 15. x² + y²  4x  6y  4 = 0, 19. 4x  3y  25 = 0 or 3x + 4y  25 = 0, , 22. (a) 3x – 4y = 21; 4x + 3y = 3; (b) A(0, 1) and B(–1, –6); (c) 90°,, , 5( 2  1) units;(d) 12.5 sq. units;, , (e) x2 + y2 + x + 5y – 6=0, x intercept 5; y intercept 7, 23. x2 + y2  2x  6y  8 = 0, 27. 9x  10y + 7 = 0, , 26. 2x  5 y 15 = 0, 2x + 5 y  15 = 0, x  35 y  30 = 0, x + 35 y  30 = 0, , 75

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JEEMAIN.GURU, , JEE-Mathematics, , EXERCISE - 04 [B], 1., , BRAIN STORMING, , SUBJECTIVE EXERCISE, , Find the equation of the circle inscribed in a triangle formed by the lines 3x + 4y = 12 ; 5x + 12y = 4 &, 8y = 15x + 10 without finding the vertices of the triangle., , 2., , Consider a curve ax2 + 2 hxy + by2 = 1 and a point P not on the curve. A line is drawn from the point P intersects, the curve at points Q & R. If the product PQ · PR is independent of the slope of the line, then show that the curve, is a circle., , 3., , Find the equation of a circle which is co-axial with circles 2x² + 2y²  2x + 6y  3 = 0 & x² + y² + 4x + 2y + 1 = 0., It is given that the centre of the circle to be determined lies on the radical axis of these two circles., , 4., , If 32 + 6 + 1 – 6m2 = 0, then find the equation of the circle for which x + my + 1 = 0 is a tangent., , 5., , Circle are drawn which are orthogonal to both the circles S  x2 + y2 – 16 = 0 and S' x2 + y2 – 8x – 12y + 16 = 0., If tangents are drawn from the centre of the variable circles to S. Then find the locus of the mid point of the chord, of contact of these tangents., , 6., , Show that the locus of the point the tangents from which to the circle x² + y²  a² = 0 include a constant angle,  is (x² + y²  2a²)² tan²  = 4a² (x² + y²  a²)., , 7., , Find the locus of the mid point of the chord of a circle x² + y² = 4 such that the segment intercepted by the, , 8., , chord on the curve x²  2x  2y = 0 subtends a right angle at the origin., Prove that the length of the common chord of the two circles x² + y² = a² and (x  c)² + y² = b² is, , 9., 10., , 11., , 1, (a  b  c) (a  b  c) (a  b  c) ( a  b  c) , where a, b, c > 0., c, Find the equation of the circles passing through the point (2, 8), touching the lines 4x  3y  24 = 0 &, 4x + 3y  42 = 0 & having x coordinate of the centre of the circle less than or equal to 8., Lines 5x + 12y  10 = 0 & 5x  12y  40 = 0 touch a circle C1 of diameter 6. If the centre of C1 lies in, the first quadrant, find the equation of the circle C2 which is concentric with C1 & cuts intercepts of length 8, on these lines., A circle touches the line y = x at a point P such that OP = 4 2 , where O is the origin. The circle contains the, point (–10,2) in its interior and the length of its chord on the line x + y = 0 is 6 2 . Determine the equation of, the circle., , [JEE 1990], , Find the intervals of values of 'a' for which the line y + x = 0 bisects two chords drawn from a point, ,  1  2a 1  2a , ,, ,  to the circle 2x2 + 2y2 – 1  2a x – 1  2a y = 0., 2,  2, , , , , 13., 14., , , , , , [JEE 1996], , Find the equations of the circles passing through (–4, 3) and touching the lines x + y = 2 and x – y = 2., P is a variable point on the circle with centre at C . CA & CB are perpendiculars from C on x-axis & y-axis, respectively. Show that the locus of the centroid of the triangle PAB is a circle with centre at the centroid of the, triangle CAB & radius equal to one third of the radius of the given circle., , ANSWER, , BRAIN STOR MIN G SUBJ ECTIVE E X ER CISE, , 1., , , , 2, , 2, , 2, , x + y  2x  2y + 1 = 0, , 5 . x 2 + y 2 – 4x – 6y = 0, , KEY, , 2, , 3. 4x + 4y + 6x + 10y  1 = 0, 7., , 1 0 . x2 + y2  10x  4y + 4 = 0, , x² + y²  2x  2y = 0, , 9., , 11. (x – 9)2 + (y – 1)2 = 50, , 1 3 . x2 + y2 + 2(10 ± 54 )x + 55 ± 8 54 = 0, , 76, , EXERCISE-4(B), , 4., , 2, , 2, , x + y – 6x + 3 = 0, , 205,  182 , , 3 , r =, centre (2,3), r = 5; centre  ,  9, , 9, 12. a  (–, –2)  (2,), , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , 12., , E

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JEEMAIN.GURU, , JEE-Mathematics, , 14., , Let C be the circle with centre (0, 0) and radius 3 units. The equation of the locus of the mid points of the chords, 2, at its centre is [AIEEE-2006, IIT-1996], 3, 27, 9, 3, (1) x2 + y2 = 1, (2) x2 + y2 =, (3) x2 + y2 =, (4) x2 + y2 =, 4, 4, 2, Consider a family of circles which are passing through the point (–1, 1) and are tangent to x-axis. If (h, k) are the, , of the circle C that subtend an angle of, , 15., , co-ordinates of the centre of the circles, then the set of values of k is given by the interval(1) 0 < k < 1/2, , (2) k  1/2, , (3) –1/2  k  1/2, 2, , [AIEEE-2007], , (4) k  1/2, , 2, , 16., , The point diametrically opposite to the point (1, 0) on the circle x + y + 2x + 4y – 3 = 0 is-, , 17., , (1) (3, –4), (2) (–3, 4), (3) (–3, –4), (4) (3, 4), Three distinct points A, B and C are given in the 2–dimensional coordinate plane such that the ratio of the, , [AIEEE-2008], , distance of any one of them from the point (1, 0) to the distance from the point (–1, 0) is equal to, the circumcentre of the triangle ABC is at the point :-, , 18., , 19., , 1, . Then, 3, , [AIEEE-2009], , 5 , 5 , 5 , (1)  , 0 , (2)  , 0 , (3) (0, 0), (4)  , 0 , 2, 3, , , , , 4 , 2, 2, If P and Q are the points of intersection of the circles x + y + 3x + 7y + 2p – 5 = 0 and, x 2 + y 2 + 2x + 2y – p 2 = 0, then there is a circle passing through P, Q and (1, 1) for :[AIEEE-2009], , (1) All except two values of p, , (2) Exactly one value of p, , (3) All values of p, , (4) All except one value of p, , For a regular polygon, let r and R be the radii of the inscribed and the circumscribed circles. A false statement, among the following is :[AIEEE-2010], (1) There is a regular polygon with, , r 1, , R 2, , r, 1, , R, 2, , (2) There is a regular polygon with, , r 2, r, 3, , , (4) There is a regular polygon with, R 3, R, 2, The circle x2 + y2 = 4x + 8y + 5 intersects the line 3x – 4y = m at two distinct points if :[AIEEE-2010], (3) There is a regular polygon with, , (1) – 85 < m < – 35, 21., 22., , (2) – 35 < m < 15, , (3) 15 < m < 65, , The two circles x 2 + y 2 = ax and x 2 + y 2 = c 2 (c>0) touch each other if :(1) a = 2c, (2) |a| = 2c, (3) 2|a| = c, , (2) x 2 + y 2 – 2x – 2y + 1 = 0, , [AIEEE-2011], , (3) x 2 + y 2 – x – y = 0, (4) x 2 + y 2 + 2x + 2y – 7 = 0, The length of the diameter of the circle which touches the x-axis at the point (1, 0) and passes through the, point (2, 3) is :, [AIEEE-2012], (1) 5/3, , 24., , [AIEEE-2011], , (4) |a| = c, , The equation of the circle passing through the points (1, 0) and (0, 1) and having the smallest radius is:, (1) x 2 + y 2 + x + y – 2 = 0, , 23., , (4) 35 < m < 85, , (2) 10/3, , (3) 3/5, , (4) 6/5, , The circle passing through (1, – 2) and touching the axis of x at (3, 0) also passes through the point :, [JEE(Main)-2013], , (1) (–5, 2), , (2) (2, –5), , (3) (5, –2), , ANSWER, , P RE VIOU S Y EARS QU E STION S, , (4) (–2, 5), , KEY, , E XE R CISE -5, , [A], , Q u e., , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , 11, , 12, , 13, , 14, , 15, , Ans, , 3, , 1, , 2, , 4, , 2, , 1, , 1, , 1, , 2, , 4, , 4, , 4, , 3, , 3, , 2, , Q u e., , 16, , 17, , 18, , 19, , 20, , 21, , 22, , 23, , 24, , Ans, , 3, , 4, , 4, , 3, , 2, , 4, , 3, , 2, , 3, , 78, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , 20., , E

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JEEMAIN.GURU, , JEE-Mathematics, , EXERCISE - 05 [B], 1., , Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius r. If PS and RQ intersect, at a point X on the circumference of the circle then 2r equals, (A), , 2., , JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS, , PQ  RS, (B), 2, , PQ  RS, , 2, , 2PQ  RS, (C), PQ  RS, , (D), ,  PQ , , 2, ,   RS , , 2, [JEE 2001 (Screening) 1], 2, 2, Let 2x + y – 3xy = 0 be the equation of a pair of tangents drawn from the origin 'O' to a circle of radius 3 with, centre in the first quadrant. If A is one of the points of contact, find the length of OA., [JEE 2001 (Mains), 5], , 3., , Find the equation of the circle which passes through the points of intersection of circles, x 2 + y 2 – 2x – 6y + 6 = 0 and x 2 + y 2 + 2x – 6y + 6 = 0 and intersects the circle, x2 + y2 + 4x + 6y + 4 = 0 orthogonally., [REE 2001 (Mains), 3], , 4., , Tangents TP and TQ are drawn from a point T to the circle x 2 + y 2 = a 2. If the point T lies on the line, px + qy = r, find the locus of centre of the circumcircle of triangle TPQ., [REE 2001 (Mains), 5], , 5., , If the tangent at the point P on the circle x2 + y2 + 6x + 6y = 2 meets the straight line 5x – 2y + 6 = 0 at a point, Q on the y-axis, then the length of PQ is, (B) 2 5, , (A) 4, , (D) 3 5, , (C) 5, , [JEE 2002 (Scr), 3], 6., , If a > 2b > 0 then the positive value of m for which y = mx – b 1  m 2 is a common tangent to, x2 + y2 = b2 and (x – a)2 + y2 = b2 is, , 7., , a 2  4b 2, 2b, , b, a  2b, a  4b, [JEE 2002 (Scr), 3], The radius of the circle, having centre at (2, 1), whose one of the chord is a diameter of the circle, x2 + y2 – 2x – 6y + 6 = 0, , (A), , 2b, , 2, , 2, , (A) 1, , (B), , (B) 2, , 2b, a  2b, , (C), , (D), , (C) 3, , (D), , 3, , 8., , Line 2x + 3y + 1 = 0 is a tangent to a circle at (1, –1). This circle is orthogonal to a circle which is drawn having, , 9., , A circle is given by x2 + (y – 1)2 = 1, another circle C touches it externally and also the x-axis, then the locus of, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , diameter as a line segment with end points (0, –1) and (– 2, 3). Find equation of circle., , E, , [JEE 2004 (Scr)], , its centre is, , 10., , (A) {(x, y) :, , x2, , (C) {(x, y) :, , x2, , [JEE 2004, 4], [JEE 2005 (Scr)], , = 4y}  {(x, y) : y  0}, = y}  {(0, y) : y  0}, , (B) {(x, y) :, , x2, , 1)2, , + (y –, , (D) {(x, y) :, , x2, , = 4y}  {(0, y) : y  0}, , = 4}  {x, y) : y  0}, , Let ABCD be a quadrilateral with area 18, with side AB parallel to the side CD and, AB = 2CD. Let AD be perpendicular to AB and CD. If a circle is drawn inside the quadrilateral ABCD touching, , 11., , all the sides, then its radius is, [JEE 2007, 3], (A) 3, (B) 2, (C) 3/2, (D) 1, Tangents are drawn from the point (17, 7) to the circle x2 + y2 = 169., Statement-1 : The tangents are mutually perpendicular., because, Statement-2 : The locus of the points from which mutually perpendicular tangents can be drawn to the given, circle is x2 + y2 = 338., (A) Statement-1 is true, statement-2 is true; statement-2 is correct explanation for statement-1., (B) Statement-1 is true, statement-2 is true; statement-2 is NOT a correct explanation for statement-1., (C) Statement-1 is true, statement-2 is false., (D) Statement-1 is false, statement-2 is true., , [JEE 2007, 3], , 79

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JEEMAIN.GURU, , JEE-Mathematics, , 12., , Consider the two curves C1 : y2 = 4x ;, , C2 : x2 + y2 – 6x + 1 = 0. Then,, , (A) C1 and C2 touch each other only at one point, (B) C1 and C2 touch each other exactly at two points, (C) C1 and C2 intersect (but do not touch) at exactly two points, (D) C1 and C2 neither intersect nor touch each other, 13., , Consider,, , [JEE 2008, 3], , L1 : 2x + 3y + p – 3 = 0 ; L2 : 2x + 3y + p + 3 = 0,, C : x2 + y2 + 6x – 10y + 30 = 0., , where p is a real number, and, , Statement-1 : If line L1 is a chord of circle C, then line L2 is not always a diameter of circle C., and, Statement-2 : If line L1 is a diameter of circle C, then line L2 is not a chord of circle C., (A) Statement-1 is True, Statement-2 is True; statement-2 is a correct explanation for statement-1, (B) Statement-1 is True, Statement-2 is True; statement-2 is NOT a correct explanation for statement-1, (C) Statement-1 is True, Statement-2 is False, (D) Statement-1 is False, Statement-2 is True, 14., , [JEE 2008, 3], , Comprehension (3 questions together):, A circle C of radius 1 is inscribed in an equilateral triangle PQR. The points of contact of C with the sides PQ,, QR, RP are D, E, F respectively. The line PQ is given by the equation, , 3 x + y – 6 = 0 and the point D is, , 3 3 3 , ,, ,  . Further, it is given that the origin and the centre of C are on the same side of the line PQ., 2 ,  2, , (ii), , The equation of circle C is, (B) (x – 2 3 )2 + (y +, , (C) (x – 3 )2 + (y + 1)2 = 1, , (D) (x – 3 )2 + (y – 1)2 = 1, , Points E and F are given by, ,  3 3, (A) , , ,,  2, 2 , , , , , 3,0, , , ,  3 3  3 1, (C)  2 , 2  ,  2 , 2 , ,  , , (iii), , (C) y =, , 16., ,  3 1, , ,, (B) ,  2, 2 , , , , , 3, (D)  ,, 2, ,  3 1, , , , 2 ,  2, , 3, 2, , ,  ,, , , 3, 0, , , , Equations of the sides RP, RQ are, (A) y =, , 15., , 1 2, ) = 1, 2, , (A) (x – 2 3 )2 + (y – 1)2 = 1, , 2, 3, , x + 1, y = –, , 2, 3, , x – 1, , (B) y =, , 3, 3, x + 1, y = –, x – 1, 2, 2, , 1, 3, , x, y = 0, , (D) y = 3 x, y = 0, , [JEE 2008, 4+4+4], , Tangent s draw n from the poi nt P(l, 8) to the circle x 2 + y 2 – 6x – 4y – 11 = 0, at the points A and B. The equation of the circumcircle of the triangle PAB is, (A) x2 + y2 + 4x – 6y + 19 = 0, , (B) x2 + y2 – 4x – 10y + 19 = 0, , (C) x2 + y2 – 2x + 6y – 29 = 0, , (D) x2 + y2 – 6x – 4y + 19 = 0, , touch the circle, , [JEE 2009, 3], , The centres of two circles C1 and C2 each of unit radius are at a distance of 6 units from each other. Let P be the, mid point of the line segment joining the centres of C1 and C2 and C be a circle touching circles C1 and C2, externally. If a common tangent to C1 and C passing through P is also a common tangent to C2 and C, then the, radius of the circle C is, [JEE 2009, 4], , 80, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , (i), , E

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JEEMAIN.GURU, 17., , JEE-Mathematics, 3  1 apart. If the chords subtend at the, , Two parallel chords of a circle of radius 2 are at a distance, , , 2, and, , where k > 0, then the value of [k] is, [JEE 10, 3], k, k, [Note : [k] denotes the largest integer less than or equal to k], The circle passing through the point (–1,0) and touching the y-axis at (0, 2) also passes through the point center, angles of, , 18., ,  3 , (A)   ,0 ,  2 , ,  5 , (B)   ,2 ,  2 , ,  3 5, (C)   , ,  2 2, , (D) (–4,0), [JEE 2011, 3, –1], , 19., , 2, , 2, , The straight line 2x – 3y = 1 divides the circular region x + y  6 into two parts. If,  3   5 3   1 1   1 1  , S   2,  ,  ,  ,  ,   ,  ,   ,,  4   2 4   4 4   8 4  , then the number of point(s) in S lying inside the smaller part is, , 20., , [JEE 2011, 4], , The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line, 2, , 2, , 4x – 5y = 20 to the circle x + y = 9 is-, , [JEE 2012, 3, –1], , 2, , 2, , (B) 20(x + y ) + 36x – 45y = 0, , 2, , 2, , (D) 36(x + y ) + 20x – 45y = 0, , (A) 20(x + y ) – 36x + 45y = 0, (C) 36(x + y ) – 20x + 45y = 0, , 2, , 2, , 2, , 2, , Paragraph for Question 21 and 22, A tangent PT is drawn to the circle x2 + y2 = 4 at the point P, 21., , [JEE 2012, 3, –1], (C) x  3y  4, , (B) y = 2, , [JEE 2012, 3, –1], (B) x  3 y  1, , (C) x  3y  1, , [JEE(Advanced) 2013, 3, (–1)], , 2, , 2, , (B) x + y – 6x + 7y + 9 = 0, , 2, , 2, , (D) x + y – 6x – 7y + 9 = 0, , (A) x + y – 6x + 8y + 9 = 0, , Node6\E_NODE6 (E)\Data\2014\Kota\JEE-Advanced\SMP\Maths\Unit#05\Eng\02 CIRCLE.p65, , (C) x + y – 6x – 8y + 9 = 0, , E, , ANSWER, , P RE VIOU S Y EARS QU E STION S, , 5., , (D) x  3y  5, , Circle(s) touching x-axis at a distance 3 from the origin and having an intercept of length 2 7 or y-axis, is (are), , 1., , (D) x  2 2y  6, , A possible equation of L is, (A) x  3y  1, , 23., , , , 3, 1 . A straight line L, perpendicular to PT is, , a tangent to the circle (x – 3)2 + y2 = 1., A common tangent of the two circles is, (A) x = 4, , 22., , , , A, C, , 2., 6., , OA = 3(3 + 10 ), A, , 7., , C, , 2, , 2, , 2, , 2, , KEY, , E XE R CISE -5, , 3., , x2 + y2 + 14x – 6y + 6 = 0;, , 8., , 2x 2, , +, , 2y2, , 4., , – 10x – 5y + 1 = 0, , 2px + 2qy = r, 9., , D, , 10. B, , 11. A, , 12. B, , 13. C, , 14. (i) D, (ii) A, (iii) D, , 15. B, , 16., , 8, , 17. 3, , 18. D, , 19. 2, , 20. A, , 22. A, , 23., , A,C, , 81, , 21. D, , [B]