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Circle, Basic Geometry with circles, 1., , Equal chords subtend equal angles at the, centre and vice-versa., , 2., , Equal chords of a circle are equidistant from, the centre and vice-versa., , 3. Angle subtended by an arc at the centre is, double the angle subtended at any point on, the remaining part of the circle, , 4., , Angles in the same segment of circle are, equal., , 5. The sum of the opposite angles of a cyclic, quadrilateral is 180° and vice-versa., ∠ A + ∠ C = ∠ B + ∠ D = 180°, , 6. , If two chords of a circle intersect either, inside or outside the circle, the rectangle, contained by the parts of one chord is equal, in area to rectangle by the parts of other., , 1., , Circle, , PA × PB = PC × PD
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7. The greater of the two chords in a circle is, nearer to the centre than lesser., ∵ AB > CD, \ p1 < p2, 8. , A chord drawn across the circular region, divides it in two parts each of which is called, a segment of the circle, 9. The tangents at the extremities of a chord of, a circle are equal., PA = PB, Circle, , Definition, , Standard Form, (x – a)2 + (y – b)2 = r2, (a, b) ≡ centre, r ≡ radius, General Equation of the circle, x2 + y2 + 2gx + 2fy + c = 0, centre ≡ (–g, –f), , Circle is defined as a locus, of a point ‘P’ which moves in, x – y plane in such a way that, its distance from the fixed point, in the same plane is always, constant., , 1, , 1, = − coefficient of x, − coefficient of y , 2, 2, , Radius =, , g 2 + f2 − c, , Example :, , Q1, Sol., Q2, , Circle, , Sol., , Find equation of circle whose radius is 3 and centre is (–1, 2)., Equation is, (x + 1)2 + (y – 2)2 = 32, Find equation of circle whose radius is 10 and centre is (–5, –6)., Equation is, (x + 5)2 + (y + 6)2 = 102, , 2.
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Necessary and sufficient condition for General, Equation of 2nd degree to represent a circle., ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, (a) Coefficient of x2 = coefficient of y2, (not necessarily unity) and, (b) Coefficient of xy = 0, Note :, The general equation of circle x2 + y2 + 2gx +, 2fy + c = 0 contains 3 independent arbitrary, constants g, f and c which means that a, unique circle passes through 3 non-collinear, points. Hence 3 points on a circle must be, given to determine the unique equation of the, circle., Nature of circle :, (x2 + y2) + 2gx + 2fy + c = 0, (i), If g2 + f2 – c > 0, ⇒ Real circle with finite radius, (ii) If g2 + f2 – c = 0, ⇒ Point circle, (iii) If g2 +f2 -c < 0, ⇒ Imaginary circle, Example :, , Sol., , Find the equation of the circle passing through the points (3, 4), (–3, –4),, (0, 5)., Let equation of circle be, x2 + y2 + 2gx + 2fy + c = 0, Now on satisfying the equation by given points,, ⇒9 + 16 + 6g + 8f + c = 0, , 6g + 8f + c = –25 , ⇒ 9 + 16 – 6g – 8f + c = 0, , 6g + 8f – c = 25 , ⇒ 0 + 25 + 0 + 10f + c = 0, , 10f + c = –25 , , …(1), …(2), …(3), , 3., , Circle, , Q1
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Subtracting (2) from (1), ⇒ 2c = -50 ⇒ c = –25, from (3) ⇒ 10f = –25 – c ⇒ f = 0, from (2) ⇒ 6g + 0 + 25 = 25 ⇒ g = 0, Hence, equation is x2 + y2 – 25 = 0, Alternate Method :, For ∆ABC, A(3, 4), B(–3, –4) , C(0, 5) origin will be circumcentre which is, centre of circle passing through A, B and C and radius r = OA = 32 + 42 = 5, equation (x – 0)2 + (y – 0)2 = 52, , Q2, Sol., , Find the equation of the circle having lines 2x – 3y = 5 and 3x – 4y = 7 as its, diameter / normal / longest chord and whose area is 154 sq. units., Centre will be point of intersection of the diameters, d1 : 2x – 3y = 5 , …(1), d2 : 3x – 4y = 7 …(2), (2) × 3 – (1) × 4, x = 1, y = –1 ⇒ centre (1, –1), Area = 154 ⇒ πr2 = 154, 154 154, r2 =, =, × 7 = 49, π, 22, Hence, equation is : (x – 1)2 + (y + 1)2 = 49, , Q3, Sol., , Find the equation of the circumcircle of ∆ formed by the lines xy + 2x + 2y +, 4 = 0, x + y + 2 = 0, xy + 2x + 2y + 4 = x(y + 2) +2 (y + 2), , = (x + 2) (y + 2), Sides of triangle L1 : x + 2 = 0, , L2 : y + 2 = 0, , L3 : x + y + 2 = 0, ∵ L 1 ⊥ L2 hence circumcentre will be mid point of hypotenuse, , (, , ), , (, , L 1 and L3 ⇒ A −2, 0 , L2 and L3 ⇒ C 0, − 2, , ), , −2 + 0 0 − 2 , Centre : P , ,, = P −1, − 1, 2 , 2, , (, , Circle, , radius =, , AC, =, 2, , ), , 4+4, = 2, 2, , (, , ) + ( y + 1), , Hence, equation is : x + 1, , 2, , 2, , =2, , 4.
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Q4, Sol., , Find centre and radius of the circle 2x2 + 2y 2 − 6x + 8y − 5 = 0 ., Given equation is, x2 + y 2 − 3x + 4y −, , , centre ≡ −, , , , 3, , 1, 1, −3 , −, 4 ≡ , −2, 2, 2, , 2, , , ( ), 2, , radius =, , 5, =0, 2, , (), , 3, − + 2, 2, , ( ), , 2, , +, , 9, 5, +4+ =, 4, 2, , Q5, Sol., , 5, 2, , 35, unit, 2, , Find equation of circle concentric with 3x2 + 3y2 – 5x – 6y – 14 = 0 and perimeter of its semicircle is 36., Given circle is x2 + y 2 −, , 5, 14, x − 2y −, =0, 3, 3, , 5 , Centre: , 1 , 6 , perimeter of semicircle, 2r + πr = 36, , 22 , r 2 +, = 36 ⇒ r = 7, 7 , , 2, , , 5, Hence, equation of required circle is x − + y − 1, 6, , , , Sol., , ), , 2, , = 49, , Find equation of the circle which passes through (2, 3) and centre on the, x-axis, radius being 5., Let centre be (α, 0), given r = 5, Equation will be (x – α)2 + (y – 0)2 = 52, which passes through (2, 3) hence, (2 – α)2 + 9 = 25, (2 – α)2 = 16 ⇒ 2 – α = 4, –4, , α = -2 , 6, Hence, equations are (x + 2)2 + y2 = 25, or (x – 6)2 + y2 = 25, , 5., , Circle, , Q6, , (
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Q7, , Find the equation of the circle for which centre is on the line y = 2x and circle passes through (–1, 2) and (3, –2)., , Sol., , Perpendicular bisector of AB where A (–1, 2) and B (3, –2) will be another diameter, Hence, mid-point will be, (1, 0), 4, slope mAB =, = −1, −4, Equation will be (y – 0) = 1 (x – 1) ⇒ y = x – 1 , …(1), and y = 2x …(2), from (1) and (2), centre O (–1, –2), Radius = OA =, , 02 + 42 = 4, , Hence, equation is : (x + 1)2 + (y + 2)2 = 42, , Q8, Sol., , 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., Let centre be O(α, β), A(0, 1), which lies on x + y = 2, ⇒ α + β = 2 …(1), Now radius is length of perpendicular from O to line 4x – 3y + 4 = 0, (as line of 2 tangent to circle), 4α − 3β + 4, Hence,, = OA (radius), 5, , (, , ), , 4α − 3β + 4 = 5 α2 + β − 1, , (, , ), , 2, , (, , ), , 4α − 3 2 − α + 4 = 5 α2 + 2 − α − 1, , 2, , 7α − 2 = 5 2α2 − 2α + 1, On squaring,, , (, , ), , 49α2 − 28α + 4 = 25 2α2 − 2α + 1, α2 − 22α + 21 = 0, , ( α − 21)( α − 1) = 0, , Circle, , ⇒ α = 1, 21, β = 1, –19, Centre will be (1, 1) or (21, –19), for centre (1, 1), radius =, , 12 + 02 = 1, 6.
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Equation is: (x – 1)2 + (y –1)2 = 1, for centre (21, –19), radius= 212 + 202 = 841 = 29, equation : (x – 21)2 + (y + 19)2 = 292, , Locus of the image of the point (2, 3) in the line, (2x – 3y + 4) + k(x – 2y + 3) = 0, k ∈ R is a, (A) Circle of radius 3 , (B) Straight line parallel to X – axis, (C) Straight line parallel to Y – axis, , Ans., Sol., , (D) Circle of radius, , 2, , (D), Given family of straight line, L 1 + kL2 = 0, will always pass-through point of intersection of L1 and L2, L 1 : 2x − 3y + 4 = 0 …(1), L2 : x − 2y + 3 = 0 …(2), (1)–(2) × 2, y = 2, x = 1 these point A(1, 2), Now let image of B(2, 3) is P(h, k) then perpendicular bisector of BP will pass, through A, h + 2 k + 3, hence mid-point M , ,, , 2 , 2, slope = mBP =, , k−3, h−2, , , h − 2, k + 3, h+ 2, Equation is: y −, = −, x −, , 2 , 2 , , k − 3, which passes through A hence, , h − 2, k + 3, h+ 2, 2 −, = −, 1 −, , 2 , 2 , , k − 3, , (, (, , ), ), , h − 2 −h , 1−k, =−, , , 2, k−3 2 , k − k2 − 3 + 3k = h2 − 2h, , ⇒ h2 + k2 − 2h − 4k + 3 = 0, Locus: x2 + y 2 − 2x − 4y + 3 = 0, centre (1, 2), radius = 2, 7., , Circle, , Q9
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Q10, Sol., , Find the equation of circle whose centre is (4, 3) and touches the line, 5x – 12y – 10 = 0., Centre C(4, 3), Radius = Perpendicular distance of C from given tangent, r=, , ( ), , ( ), , 5 4 − 12 3 − 10, , =2, , 13, , Equation is :, , ( x − 4 ) + ( y − 3), 2, , 2, , =4, , Diametrical Form of Circle, , (x − x ) (x − x ) + ( y − y ) ( y − y ) = 0, 1, , 2, , 1, , 2, , Where (x1, y1) and (x2, y2) are diametrical, opposite ends., Examples :, , Q1, Sol., Q2, Sol., , Find the equation of the circle of least radius passing through the points., (2, 3), (3, 1)., Given point will be diametrical end points of required circle, Equation: (x – 2) (x – 3)+(y – 3)(y – 1) = 0, (x2 + y2) – 5x – 4y + 9 = 0, Find equation of tangent to circle parallel to tangent x + y = 5, where centre, of the circle is (1, 2)., Let tangent is x + y = λ, Since both are tangents hence perpendicular distance of tangent from centre, will be radius, 1+2− λ, 1+2−5, ⇒, =, 2, 2, ⇒ 3 − λ = 2 ⇒ λ = 5, 1, , Circle, , Equation: x + y = 1, , 8.
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Q3, , The abscissa of 2 points ‘A’ and ‘B’ are the roots of the equation, x2 + 2x − 3 = 0 and the ordinate are the roots of the equation y 2 − 4y + 1 = 0 ., Find the equation of circle AB as diameter., , Sol., , (, , Let A x 1 , y 1, , ), , (, , and B x2 , y 2, , ), , Now given, x 1 , x2 are roots of x2 + 2x − 3 = 0, , (, , ⇒ x2 − 2x − 3 = x − x 1, , )( x − x2 ) …(1), , y 1 , y 2 are roots of y 2 − 4y + 1 = 0, , (, , ⇒ y 2 − 4y + 1 = y − y 1, , )( y − y2 ) …(2), , Equation of circle diameter AB is, x − x 1 x − x2 + y − y 1 y − y 2 = 0, , (, , )(, , ) (, , )(, , ), , x2 + y 2 + 2x − 4y − 2 = 0 , , Sol., , Find the equation of the circle which touches the line x = 0, y = 0 and x = 4?, Let radius = r, Now, Diameter = distance between x = 0, x = 4, 2r = 4 ⇒ r = 2, 0 + 4, , Centre will be , , ± r ≡ 2, 2 or 2, − 2, 2, , , (, , (, , Equation : x − 2, , Q5, Sol., , ) + ( y ± 2), 2, , 2, , ), , (, , ), , = 22, , Line y = mx + c cuts the curve y2 = 4ax at A and B. Find the equation of circle, with AB as diameter., , (, , ), , (, , Let A x 1 , y 1 andB x2 , y 2, , ), , y = mx + c , …(1), and y2 = 4ax …(2), by (1) and (2),, x1, (mx + c)2 = 4ax, , x2, , m2x2 + (2mc – 4a)x + c2 = m2(x – x1)(x – x2) …(3)2, Again by (1) and (2), 9., , Circle, , Q4, , {from (1) and (2)}
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y − c, y 2 = 4a , , m , , y1, ⇒my – 4ay + 4ac = 0, 2, , y2, , my – 4ay + 4ac = m(y – y1)(y – y2) …(4), Equation: m2 (x – x1)(x – x2) + m2(y – y1)(y – y2) = 0, m2x2 + m2y2 + (2mc – 4a)x – 4amy + c2 + 4acm = 0, 2, , Q6, Sol., , The line lx + 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)= l2 + m2., Equation of pair of Straight lines by homogenization, lx + my , ax + 2hxy + by = , , −n, , , 2, , 2, , (n a − l ) x + (2hn, 2, , 2, , 2, , 2, , ), , 2, , (, , ), , − 2lm xy + n2b − m2 y 2 = 0, , Which must be perpendicular, (∵ circle with PQ diameter passes through origin), ⇒ coeff. of x2 + coeff. of y 2 = 0, , (n a − l ) + (n b − m ) = 0, 2, , (, , 2, , 2, , 2, , ), , n2 a + b = l 2 + m2 Hence proved., , Q7, , Circle, , Ans., Sol., , A rectangle is inscribed in a circle with a diameter lying along the line,, 3y = x + 7. If the two adjacent vertices of the rectangle are (–8, 5) and (6, 5), then the area of the rectangle (in sq. units) is:, (A) 72 (B) 84 (C) 98 (D) 56, (B), Diameter, L : x – 3y + 7 = 0 , …(1), Given points A(–8, 5) and B(6, 5) lies on same side of line L, Perpendicular bisector of AB will be another diameter, mid point of AB : M(–1, 5), mAB = 0, equation of perpendicular bisector AB : x = –1, …(2), On solving (1) and (2), centre O(–1, 2), O is mid point of BD, hence D(–8, –1), ar(ABCD) = AB × AD, = 14 × 6 = 84 sq. unit, 10.
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Q8, Sol., , If line y = x + c and y2 = 8x intersect in A and B. Circle with AB as diameter, passes through (0, 0). Find c?, By homogenization equation of pair of straight line, y −x, y2 = 8x , , c , ⇒ 8x2 + cy 2 − 8xy = 0, Lines are perpendicular (∵ circle with diameter passes through origin), ⇒ 8 + c = 0 ⇒ c = –8, , Q9, Sol., , Find locus of point of intersection of x + 2y + λ(x – 2y) = 0 and, (x + y – 2) + λ(x – 2) = 0 if these lines are always perpendicular to each other., First family of straight lines L1 + λL2 = 0 passes through, point of intersection of, L1 : x + 2y = 0 and L2 : x – 2y = 0 which is A(0, 0), Second family of straight lines L3 + λL4=0 passes, through point of intersection of, L3 : x + y = 2 and L4 : x –2 = 0 which is B(2, 0), Now required locus will be circle as lines from A and, B are perpendicular and A and B will be diametric end, points, Equation : (x – 0)(x – 2) + (y – 0)(y – 0) = 0, , x2 + y2 –2x = 0, , INTERCEPT, Length of chord:, AB = 2MB, , 11., , Circle, , AB = 2 r2 − p2
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Angle between Line and circle:, , cosθ =, , p, r, , X-intercept:, , x 1 − x2 = 2 g 2 − c, Proof :, , , AB = 2MB, , , , = 2 r2 − p2, , , , =2, , , , = 2 g2 − c, , (g, , 2, , ) ( ), , + f 2 − c − −f, , 2, , (i), , If g2 > c, ⇒ Circle cuts the x-axis at 2 distinct points, (ii) If g2 = c, ⇒ Circle touches x-axis, (iii) If g2 < c, ⇒ Circle lies completely above or below the, x-axis, Y-intercept :, , y 1 − y2 = 2 f2 − c, Proof :, , AB = 2MB, , , = 2 r2 − p2, , , , =2, , , , = 2 f2 − c, , (i), , If f2 > c, , 2, , ) ( ), , + f 2 − c − −g, , 2, , ⇒ circle cuts the y-axis at 2 distinct points, , 2, , ⇒ circle touches y-axis, , 2, , ⇒ circle lies completely either on right or on left of y-axis, , (ii) If f = c, (iii) If f < c, Circle, , (g, , 12.
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Examples :, , Sol., , Q2, , Sol., Q3, Sol., , Find the equation of the circle which touches, the positive axis of y at a distance of 4 units, from origin and cuts off an intercept of 6 unit, from the positive axis., , PQ = 6, In ∆PCM, rỐ = 3 + 4 ⇒ r = 5, Centre : C(r, 4) ≡ C(5, 4), Equation : (x – 5)2 + (y – 4)2 = 52, Find the equation of circle which touches, the co-ordinate axes and whose radius = 5., , Equation:, , ( x ± 5) + ( y ± 5), 2, , 2, , = 52, , Find the equation of a circle through origin cutting off intercept equals to, unity on the lines y2 – x2 = 0., , (, , )(, , y 2 − x2 = y + x y − x, , ), , L1 : x + y = 0, , L2 : x − y = 0, 1, 1 1, 1 , A , ,, ,−, ,B , , 2, 2 2 2, −1 1 , −1, 1 , C , ,, ,−, and D , , 2, 2 2, 2, 13., , Circle, , Q1
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for required circles AB, AC, CD, BD are diameters, , 1 , 1 , 1 , 1 , Equation: x ±, x ±, + y ±, y ±, = 0, 2 , 2 , 2 , 2, , 2, , 2, , , , 1 , 1 , + y ±, = 0, x ±, 2, 2, , , , ⇒ x2 + y 2 ± 2x ± 2y + 1 = 0, , Q4, Sol., , Find the equation of the locus of the centre of a circle which touches the, positive y-axis and having intercept on x-axis equals to 2l., Let centre be P(h, k), In ∆APM, ỐỐ = + …(1), Also circle touches y-axis, ∴ QP = AP, h = r ⇒ r2 = h2 …(2), from (1) and (2),, h2 = l 2 + k2 ⇒ h2 − k2 = l 2, locus : ỐỐ −, , Q5, , Circle, , Sol., , =, , Find the equation of incircle and circumcircle of the quadrilateral formed by, the lines x = 0 and y = 0, x = 16, y = 16., Centre of incircle S1 and circumcircle, S2 will be mid-point of OA, ∴ centre C(8, 8), For incircle, radius r1 = 8, S1 : (x – 8)2 + (y – 8)2=82, for circumcircle OA is diameter, S2 : (x – 0) (x – 16) + (y – 0) (y – 16) = 0, , 14.
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Q9, , If y = mx be the equation of a chord of a circle whose radius is ‘a’, the origin, of co-ordinates being one extremity of the chord and the axis of x being a, diameter of the circle. Prove that the equation of a circle of which this chord, , (, , is the diameter is 1 + m2, , Sol., , )( x, , 2, , ), , (, , ), , + y 2 − 2a x + my = 0, , Equation of given circle, x x − 2a + y 2 = 0, , (, , ), , x2 + y 2 − 2ax = 0 …(1), Chord, y = mx , from (1) and (2), xỐ + m x − 2ax = 0, , {(, , }, , ), , x 1 + m2 x − 2a = 0 ⇒ x 1 =, , …(2), , 2a, 1 + m2, , 2am, , ∵ y 1 = mx 1 ⇒ y 1 =, , 1 + m2, for required circle OA is diameter hence, Equation: x x − x 1 + y y − y 1 = 0, , (, , x2 + y 2 −, , 2ax, 2, , 1+ m, , ( 1 + m )( x, 2, , ), , 2, , ), , −, , (, , 2amy, 1 + m2, , (, , ), , =0, , ), , + y 2 − 2a x + my = 0 Hence proved., , Q10 Prove that the equation to the circle of which the points (x , y ) and (x , y ), 1, , 1, , 2, , 2, , are the ends of a chord of a segment containing an angle θ, is, (x1 – x1)(x – x2) + (y – y1)(y –y2) ± cot θ [(x – x1)(y – y2) – (x – x2)(y – y1)] = 0, , Sol., , mAP =, mBP =, , y – y1, , x – x1, y − y2, , x − x2, , Now angle between AP and BP is θ, hence, m − mBP , tanθ = ± AP, 1 + m m , AP BP , , , Circle, , ± tanθ =, , y − y1, y − y2, −, x − x1, x − x2, , y − y 1 y − y2 , 1+ , x − x x − x , 1 , 2 , , 16.
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Q5, Sol., , If the variable line 3x – 4y +k = 0 lies between the circles x2 + y2 –2x – 2y +, 1 = 0 and x2 + y2 – 16x – 2y + 61 = 0 without intersecting or touching either, circle, then the range of k is (a, b) where a, b∈ I. Find the value of (b – a)., For first circle, C1(1, 1), r1 = 1, for second circle, C2(8, 1), r2 = 2, Clearly p1 > r1, 3−4+ k, > 1⇒ k−1 >5, 5, , (, , ) (, , ), , k ∈ −∞, −4 ∪ 6, ∞ …(1), , p2 > r2, 24 − 4 + k, > 2 ⇒ 20 + k > 10, 5, , (, , ) (, , ), , k ∈ −∞, −30 ∪ −10, ∞ …(2), , also C1 and C2 lies opposite side of line, (3 – 4 + k) (24 – 4 + k) <0, (k – 1) (k + 20) < 0 ⇒ k ∈ (–20, 1) , (1) ∩ (2) ∩ (3), ⇒ k ∈ (–10, –4), hence a = –10, b = –4, b – a = 10 – 4 = 6, , …(3), , Line and a Circle, Let L = 0 be a line and S = 0 be a circle. If ‘r’, is the radius of the circle and ‘p’ is the length, of perpendicular from the centre on the line,, then, , II Method, Solve, (i) D > 0, (ii) D = 0, (iii) D < 0, , the line with the circle and if, ⇒ line is Secant., ⇒ line is Tangent., ⇒ line passes outside the circle., , 19., , Circle, , Line and a Circle, (i) If p > r ⇒ line is neither secant nor tangent, and passes outside the circle., (ii) If p = r ⇒ line is tangent to the circle., (iii) If p < r ⇒ line is secant., (iv) If p = 0 ⇒ line is a diameter.
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Example:, , Q1, Sol., , For what value of ‘m’ the line 3x – my + 6 = 0 is tangent to the circle, x2 + y2 – 4x + 6y – 3 = 0., Centre C(2, –3), radius = r =, , Now p = r, ⇒, , ( ), , 4+9+3, , r=4, , ( ), , 3 2 − m −3 + 6, 9 + m2, , =4, , ⇒ 12 + 3m = 4 9 + m2, On squaring,, 144 + 9m2 + 72m = 144 + 16m2, ⇒ 7m2 – 72m = 0, 72, ⇒ m = 0 or m =, 7, , Q2, , The sum of square of lengths of the chords intercepted on the circle x2 + y2 = 16,, by the lines, x + y = n, n ∈ N where N is the set of all natural numbers, is, (A) 320 (B) 160 (C) 105 (D) 210, , Ans. (D), Sol. Centre C(0, 0), radius r = 4, p=, , 0+0− n, 2, , ⇒p=, , n, , 2, , Length of chord = 2 r2 − p2, , , = 2 16 −, , n2, = 64 − 2n2, 2, , where n can be 1, 2, 3, 4, 5, sum of squares of length of chords, 5, , =, , ∑ (64 − 2n ), 2, , r=1, , Circle, , 5 × 6 × 11 , = 64 × 5 − 2 , = 42 × 5 = 210, 6, , , 20.
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Note :, (i) Number of tangents from external point to, circle is 2., (ii) Number of tangents from interior point to, circle is 0., (iii) If point is on periphery then number of, tangent is 1., Parametric Equation of a circle, x = x 1 + rcosθ and y = y 1 + rsinθ, (x1 , y1) → fixed centre,, r → fixed radius and, θ ∈ [0, 2π) is a parameter., Note :, If θ is eliminated we get cartesian form of a, 2, 2, circle i.e. x − x 1 + y − y 1 = r2, , (, , ) (, , ), , Example :, , Q1, Sol., , Convert x2 + y2 – 6x + 4y – 3 = 0 in parametric form., , (, , ) (, , Centre 3, −2 ≡ x 1 , y 1, , ), , r = 9 + 4 + 3 ⇒ r = 4, parametric form, x = x 1 + rcosθ, y = y 1 + rsinθ, x = 3 + 4cosθ,, , Q2, Sol., , y = −2 + 4sinθ, , If x2 + y2 – 2x – 4y – 4 = 0, find maximum and minimum value of 3x + 4y., Centre C(1, 2), radius r = 3, Parametric form: x = 1 + 3 cosθ, y = 2 + 3sinθ, 3x + 4y = 11 + (9cosθ + 12sinθ), 11 − 92 + 122 ≤ 3x + 4y ≤ 11 + 92 + 122, , Circle, , Min. value = 11 – 15 = – 4, Max. value = 11 + 15 = 26, 22.
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Parametric form :, x 1 = rcosθ, , 0, 2π, y 1 = rsinθ , , ), , Equation of tangent is, xcosθ + ysinθ = r, Proof :, Equation using Cartesian form, , x(r cosθ) + y(r sinθ) = r2, ⇒, x cosθ + y sinθ = r, Slope Form :, x2 + y 2 = a 2, Equation of the tangent is, y = mx ± a 1 + m2, , Proof :, Let equation of tangent is y = mx + c, Now p = r, , ( ), , m 0 −0+ c, 2, , m +1, , = a ⇒ c = ±a 1 + m2, , Equation: y = mx ± a 1 + m2 Hence proved., Note :, For a unique value of m there will be 2, tangent which are parallel to each other., Point of Tangency, Method-I, Step-1:, , Write equation of normal {Perpendicular to, T = 0 and passing through (-g, -f)}, Step-2:, Intersection of N = 0, T = 0 is co-ordinate, of that point., Method-II, , 27., , Circle, , Compare with point form.
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Example :, , Q1, Sol., , Find point of tangency if equation of tangent is 3x + 4y = 50 to the circle, x2 + y2 – 6x–8y = 0, Equation of Normal is, 4x – 3y = λ, It passes through centre (3, 4) hence, 4(3) – 3(4) = λ ⇒ λ = 0, Normal : 4x = 3y, Tangent : 3x + 4y = 50, 4x , On solving, 3x + 4 , = 50 ⇒ 25x = 150, 3 , x = 6, y = 8, Point of tangency : (6, 8), , Q2, Sol., , A circle passes through the points (-1, 1), (0, 6) and (5, 5). Find the points on, this circle, the tangent at which are parallel to the straight line joining the, origin to its centre., Let A(–1, 1), B(0, 6) and C(5, 5), AB= 52 + 12 = 26, BC= 52 + 12 = 26, AC= 62 + 42 = 52, ∵ AC2 = AB2 + BC2 ⇒ ∆ ABC is right angled, ∴ Circumcentre (mid-point of AC) is P(2, 3), 32 + 22 =, , radius = r = PA =, , (, , equation : x − 2, , ) + ( y − 3), 2, , 2, , 13, = 13 …(1), , x2 + y 2 − 4x − 6y = 0, Slope of line joining O and P: mop =, , 3, 2, , ) −32 ( x − 2) …(2), 2, 2, 4, by (1) and (2), ( x − 2 ) + ( x − 2 ) = 13, 9, 2, ( x − 2) = 9 ⇒ x = 5, −1, (, , equation of normal: y − 3 =, , Circle, , y = 1, 5, , required points are (5, 1) and (–1, 5), , 28.
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Q3, Sol., , Find the equation of the tangents to the circle x2 + y2 – 2x – 4y – 4 = 0, which are, (i) Perpendicular to the line 3x – 4y + 7 = 0., Centre (1, 2), radius r = 1 + 4 + 4 = 3, Let tangent is 4x + 3y + k = 0, Now p = r, , (), , ( ), , 4 1 +3 2 + k, 5, , = 3 ⇒ 10 + k = 15, , K = 5, –25, Equation: 4x + 3y + 5 = 0, 4x + 3y – 25 = 0., (ii) Parallel to the line 3x – 4y + 7 = 0., , Sol., , Let tangent be 3x − 4y + k = 0, now p = r, , (), , ( ), , 3 1 −4 2 + k, 5, , =3, , k − 5 = 15 ⇒ k = 20, −10, , equation: 3x – 4y + 20 = 0, 3x – 4y – 10 = 0, , Sol., , Find the equation of the tangent to the circle x2 + y2 = 4 drawn from the point, (2, 3)., Centre O(0, 0), radius r = 2, Let tangent be (y – 3) = m(x – 2), mx – y + 3 – 2m = 0, Now P = r, , ( ), , m 0 − 0 + 3 − 2m, m2 + 1, , =2, , 3 − 2m = 2 m2 + 1, , 29., , Circle, , Q4
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on squaring,, 9 + 4m2 –12m = 4m2 + 4, 5, ⇒ m=, 12, Clearly given point (2, 3) is outside the circle hence two tangents can be drawn, 5, in which one tangent have slope, and other must be parallel to y-axis, 12, , (, , ), , ∴ equation : x= 2, y − 3 =, , Q5, , Circle, , Ans:, Sol., , 5, x−2, 12, , (, , ), , A circle touches the y-axis at the point (0, 4) and passes through the point (2,, 0). Which of the following lines is not a tangent to this circle?, (A) 4x – 3y + 17 = 0 , (B) 3x + 4y – 6 = 0, (C) 4x + 3y – 8 = 0 , (D) 3x – 4y – 24 = 0, (C), AC = r, ⇒ AC2 = r2, (r – 2)2 + 42 = r2, r2– 4r + 4 + 16 = r2, 4r = 20, r=5, equation of circle :, (x – 5)2 + (y – 4)2 = 52, centre C(5, 4), r = 5, Now check p = r for options, 20 − 12 + 17, (A) p =, = 5 = r , 5, (B) p =, , 15 + 16 − 6, = 5 = r , 5, , (C) p =, , 20 + 12 − 8, 24, =, ≠r, 5, 5, , (D) p =, , 15 − 16 − 24, = 5 = r , 5, , ∵ , , , 30.
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Sol., , Find the equation of the tangent drawn to the circle x2 + y2 – 6x + 4y – 3 = 0, from the point (7, 4) lying outside the circle. Also find the point of contact., Centre (3, –2), radius r = 9 + 4 + 3 = 4, Circle S : (x – 3)2 + (y + 2)2 = 16 …(1), equation of tangent from slope from, , ( y + 2) = m ( x − 3), , ± r 1 + m2, , ( y + 2) = m ( x − 3), , ± 4 1 + m2, , Which passes through (7, 4) hence, 6 = 4m ± 4 1 + m2, , ( 3 − 2m), , 2, , (, , = 4 1 + m2, , ), , 9 − 12m + 4m2 = 4 + 4m2, , ⇒ m=, , 5, and m = Not defined, 12, , 5, 13, x−3 ±, , x=7, 12, 3, but (–) sign tangent does not-satisfy point (7, 4), 5, 13, Hence, equation T1 : y + 2 =, x−3 +, …(2), 12, 3, , (, , ), , (, , equation y + 2 =, , (, , ), , ), , (, , ), , T2 : x = 7 …(3), for point of contact, for T1 : 5x − 12y + 13 = 0, N1 : 12x + 5y − 26 = 0, 95 22 , on solving point of contact is , ,, , 13 13 , for T2 : x = 7, , (, , put in equation of circle 42 + y + 2, , ), , 2, , = 42 ⇒ y = −2, , ∴ point (7, –2), , 31., , Circle, , Q6
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Q7, Sol., , Find shortest distance between line 3x + 4y = 25 and circle x2 + y2 – 6x + 8y = 0., Centre C(3, –4), radius r = 9 + 16 = 5, p=, , ( ), , ( ), , 3 3 + 4 −4 − 25, 5, , 32, 5, Shortest Distance = AB = p – r, 32, 7, =, −5 =, unit, 5, 5, p=, , Q8, Sol., , If equation of tangent line on circle x2 + y2 = 1 is y = x + 2 then find point of, contact., Let point be P(x1 , y1) then tangent, xx1 + yy1 = 1 …(1), Also given equation of tangent, x − y = − 2 …(2), , By comparing (1) and (2),, x1, y, 1, = 1 =, 1, −1 − 2, ⇒ x1 = −, , 1, 2, , , y1 =, , 1, 2, , −1 1 , point is , ,, , 2 2, , Q9, Sol., , Tangent is drawn from the point P(4, 0) to the circle x2 + y2 = 8 touches it at, the point A in the 1st quadrant. Find the co-ordinates of another point B on the, circle such that AB = 4., OA = 2 2, OP = 4, ⇒ AP = 2 2, , Circle, , hence ∠AOP =, , π, 4, , 32.
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, π, π , hence A 2 2 cos , 2 2 sin = A 2, 2, 4, 4 , , By diagram we can see, , , 3π, 3π , 3π, 3π , B 2 2 cos, , 2 2 sin, , −2 2 sin, or B −2 2 cos, , 4, 4 , 4, 4 , , , , ( ), , ⇒ B(–2, 2) or B(2, –2), , Q10, Sol., , Find the locus of the middle points of portions of tangents to the circle, x2 + y2=a2 terminated by the co-ordinate axes., Let mid-point be P(h, k) and equation of tangent is, x cos θ + y sin θ = a …(1), a, , , a , which meets x-axis at A , , 0 , B 0,, , cosθ , sinθ , now P is mid-point of AB hence, a, b, = h,, =k, 2cosθ, 2sinθ, 2, , 2, , a , b , a2 b2, + = 1⇒ 2 + 2 = 4, h, k, 2h , 2k , Locus:, , x2, , +, , b2, y2, , =4, , Let RS be the diameter of the circle x2 + y2 = 1 where S is the point (1, 0). Let, P be a variable point (other than R and S) on the circle and tangents to the, circle at S and P meet at the point Q. The normal to the circle at P intersects, a line drawn through Q parallel to RS at point E. Then the locus of E passes, through the point(s), 1 1 , 1 1, (A) ,, (B) , , 3 3 , 4 2, , , , Ans., Sol., , 1 −1 , (C) ,, 3 3 , , , , 1 −1 , (D) , , 4 2 , , (AC), Let P (cosθ, sinθ), equation of tangent at P:, xcosθ + ysinθ = 1 …(1), equation of tangent at S(1, 0), x = 1 …(2), from (1) and (2), 33., , Circle, , Q11, , a2
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Example :, , Q1, Sol., , Tangents PA and PB are drawn from P(4, 3) to circle x2 + y2 = 32. Find, (a) PA, PA = S1, = 42 + 32 − 32 = 4, ⇒ L = PA = 4, , (b) Area of quadrilateral PAOB, , Sol., , Area = rL, = 3 × 4 = 12 sq. units., (c) AB, , Sol., , AB =, , 2rL, 2, , =, , 2, , ( )( ), , 2 3 4, , r +L, 32 + 42, 24, sq. unit., =, 5, , (d) area of ∆PAB, , Sol., , (, , ), , ar ∆PAB =, , rL3, 2, , 2, , r +L, , =, , ( ), , 3 4, , 3, , 9 + 16, , =, , 192, sq. unit, 25, , (e) ∠APB, 2rL , 2θ = tan−1 2, 2 , L − r , , ( )( ) = tan−1 24 , , , (, ) , 7 , , 2 3 4, = tan−1 , 16 − 9, , , 39., , Circle, , Sol.
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(f) equation of circumcircle of ∆PAB, , Sol., Q2, Sol., , endpoints of diameter are P(4, 3), O(0,0), Equation: (x–4) (x–0) + (y–3) (y–0) = 0, x2 + y2 – 4x – 3y = 0, Find the length of the tangent from any point on the circle x2 + y2 = 25 to the, circle x2 + y2 = 16., Let point on first circle be P(x1, y1), hence x21 + y 21 = 25 …(1), Length of tangent, , = S2, x21 + y 21 − 16, , , , =, , , , = 25 − 16 = 3 unit, , Alternative method : = r12 − r22, = 25 − 16 = 3 unit, , Q3, Sol., , Find the range of ‘p’ for which the power of a point P(2, 5) is negative w.r.t. a, circle x2 + y2 – 8x – 12y + p = 0 and the circle neither touches nor intersects, the co-ordinate axis., Centre C(4, 6), radius r =, , , 16 + 36 − p, , r = 52 − p, , ∵ P point lies inside the circle, ∵ S1 < 0, 22 + 52 – 16 – 60 + p < 0, ⇒ p < 47 ...(1), also circle neither intersects nor touches co-ordinate axis hence, r<4, 52 − p < 4 ⇒ 52 − p < 16, , by (1) ∩ (2),, , p > 36 , , ...(2), , Circle, , 36 < p < 47, , 40.
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Example :, , Q1, Sol., , Find the range of ‘a’ such that the angle ‘θ’ between the pair of tangents, π, <θ<π, drawn from the point (a, 0) to the circle x2 + y2 = 4 satisfies, 2, for θ =, , π, , P lies on director circle, 2, , ⇒ OP = 2 2, also for θ = π, OP = 2, hence 2 < OP < 2 2, , , 2< a <2 2, , (, , ) (, , ⇒ a ∈ −2 2, − 2 ∪ 2, 2 2, , ), , Chord in terms of Mid-Point, T = S1, Examples :, , Q1, Sol., , Find mid points of chord 2x – 5y + 18 = 0 of the circle x2 + y2 – 6x + 2y – 54 = 0, Centre C(3, –1), Midpoint will be foot of perpendicular from, centre on chord hence,, 2 3 − 5 −1 + 18 , x−3 y+1, =, = −1 , , −5, 2, 4 + 25, , , , ( ), , ( ), , Circle, , x−3 y+1, =, = −1 ⇒ x = 1, y = 4, 2, −5, midpoint M(1, 4), , 42.
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Q2, Sol., , Locus of the middle point of the chords of the circles x2 + y2 + 2gx + 2fy + c = 0, which passes through a fixed point (a, b) lying outside the circle., Let mid point be M(h, k), equation of chord : T = S1, xh + yk + g x + h + f y + k + c = h2 + k2 + 2gh + 2fk + c, , (, , ), , (, , ), , Also this chord passes through (a, b) hence, ah + bk + g(a + h) + f(b + k) = h2 + k2 + 2gh + 2fy, locus : ax + by + ag + fb = x2 + y2 + gx + fy, , Sol., , Find the equation to the locus of the middle point of the chord of the circle, x2 + y2 + 2gx + 2fy + c = 0 which subtends right angle at a given point (a, b)., for ∆AMN, P is circumcentre, ⇒ AP = r, and CM2 = CP2 + MP2, R2 = CP2 + r2, , (, , ) + (k + f ) + (h − a ) + (k − b ), , (, , ) + ( y + f ) + ( x − a ) + ( y − b), , g 2 + f2 − c = h + g, locus:, , g 2 + f2 − c = x + g, , Q4, Sol., , 2, , 2, , 2, , 2, , 2, , 2, , 2, , 2, , Tangents are drawn to a unit circle with centre at origin from every point on, the line 2x + y = 4. Prove that, (i) chord of contact passes through a fixed point, Circle S : x2 + y2 = 1, let point on given line P(h, k), ⇒ 2h + k = 4 , ….(1), equation of chord of contact w.r.t. P, xh + yk = 1 , …..(2), by (1) and (2), 1 1, x y, 1, = = ⇒ x, y ≡ , , 2, 1 4, 2 4, , (, , ), , 1 1, ∴ chord of contact passes through fixed point , , 2 4, 43., , Circle, , Q3
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L:y=0, which always passes through point of intersection of S = 0 and L = 0, now intersection x2 + 02 – 2x – 8 = 0, , (x – 4)(x + 2) = 0 ⇒x = –2, 4, points A(–2, 0) and B(4, 0), Let P(x1, y1), hence x1 + 2y1 + 5 = 0, ……(2), now equation of chord of contact AB from P, , (, , ), , , , , , xx 1 + yy 1 − x + x 1 +, , λ, y + y1 − 8 = 0, 2, , (, , ), , ( x1 − 1) x + y 1 + λ2 y − x1 + λ2 y 1 − 8 = 0, , , , on comparing it with y = 0, x 1 = 1,, , x1 −, , λ, y +8=0, 2 1, , λ, 18, y1 ⇒ y1 =, 2, λ, in (2),, 9=, , 18 , −36, ⇒ λ = −6, 1 + 2 + 5 = 0 ⇒ 6 =, λ, λ , 2, 2, equation of circle : x + y − 2x − 6y − 8 = 0, , Sol., , Find the equation of a circle which passes through the point of contact of the, tangents drawn from the origin to the circle x2 + y2 –11x + 13y + 17 = 0., Equation of chord of contact from origin, x + 0, y + 0, L : x 0 + y 0 − 11 , + 13 , + 17 = 0, 2 , 2 , , ( ), , L:, , ( ), , −11x 13y, +, + 17 = 0, 2, 2, , L : 11x – 13y –34 = 0, required family of circle is S + λL = 0, , (x, , 2, , ) (, , ), , + y 2 − 11x + 13y + 17 + λ 11x − 13y − 34 = 0, , 49., , Circle, , Q3
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Type-3 :, Equation of the family of circles passes, through two given points A(x1, y1) and B(x2, y2), Equation of circle passing through, (x1, y1), (x2, y2) in diametrical form :, S + λL = 0, , (, , S ≡ x − x1, , )( x − x2 ) + ( y − y 1 )( y − y2 ) = 0, , L ≡ line in two point form, Example :, , Q1, , Find equation of circumcircle of ∆ whose vertices are (1, 0), (2, 0), (3, 1), , Sol., , Let A(1, 0), B(2, 0), C(3,1), Equation of circle passing through A and B, (x–1) (x–2) + y2 + λ (y) = 0, which also passes through C hence., (3–1) (3–2) + 12 + λ (1) = 0 ⇒ λ =–3, Equation: (x–1) (x–2) + y2 – 3y = 0, x2 + y2 – 3x – 3y + 2 = 0, Type–4 (Point circle), Equation of family of circles touching a line, L = 0 at its fixed point (x1, y1) is, , ( x − x1 ) + ( y − y 1 ), 2, , 2, , + λL = 0, , Examples :, , Q1, Sol., , Find the equation of a circle which touches the line 2x–y=4 at the point (1, –2), and passes through (3, 4)., , (, , ) + ( y + 2), , Equation: x − 1, , 2, , 2, , (, , ), , + λ 2x − y − 4 = 0, , also it passes through (3, 4) hence, , (, , ), , Circle, , 22 + 62 + λ 6 − 4 − 4 = 0, , 50.
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Q4, Sol., , The line 2x – 3y + 1 = 0 is tangent to a circle S = 0 at (1, 1). If the radius of the, circle is 13 . Find the equation of the circle S., Equation is, , ( x − 1) + ( y − 1) + λ ( 2x − 3y + 1) = 0, ⇒ x2 + y 2 + ( 2λ − 2 ) x − ( 2 + 3λ ) y + ( λ + 2 ) = 0, 2, , r=, , 2, , 13 ⇒ r2 = 13, 2, , 2, , 2λ − 2 , 2 + 3λ , , +, − λ + 2 = 13, 2 , 2 , , (, , ), , 13λ2 = 13 × 4 ⇒ λ = ±2, λ = 2 ⇒ x2 + y 2 + 2x − 8y + 4 = 0, λ = −2 ⇒ x2 + y 2 − 6x + 4y = 0, Type–5 :, Equation of circle passing through points of, intersection of lines, 1 = 0, 2 = 0, 3 = 0, (circumcircle of the triangle thus formed) is, given by, 1 2 + λ 2 3 + µ 3 1 = 0, , Note :, To find λ and µ, coefficient of x2 = coefficient of y2 and, coefficient of xy = 0, Type–6 :, Equation of a circle circumscribing a, quadrilateral whose sides in order are, represented the line, 1 = 0; 2 = 0; 3 = 0; 4 = 0 is given by, , Circle, , 1 3 + λ 2 4 = 0, , 52.
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Note :, To find λ, coefficient of x2 = coefficient of y2 and, coefficient of xy = 0, Pole and Polar, If through a point P in the plane of the circle,, there be drawn any straight line to meet the, circle in Q and R, the locus of the point of, intersection of the tangents at Q and R is, called the polar of the point P and Q is called, the pole of the polar., The equation to the polar of a point P(x1, y1), w.r.t. the circle x2 + y2 = a2 is given by, xx1 + yy1 = a2 and if the circle is general then, the equation to the polar becomes, xx 1 + yy 1 + g x + x 1 + f y + y 1 + c = 0, , (, , ), , (, , ), , Note :, (i) , The point (x1, y1) be on the circle then the, chord of contact, tangent and polar will be, represented by same equation., (ii) Pole of a given line Ax + By + c = 0 w.r.t. any, −Aa2 −Ba2 , circle x2 + y2 = a2 is , ,, , c, c , , (iii) If the polar of a point P pass through a point, Q, then the polar of Q passes through P., Examples :, , Sol., , Find the pole of a given line lx + my = n w.r.t. a circle x2 + y2 = a2., Let Pole be (x1, y1) then polar, xx1 + yy1 = a2 ...(1), also given equation of polar, lx + my = n ...(2), comparison of (1) and (2), a2 a2m , x1, y, a2, ,, = 1 =, ⇒ x1 , y 1 ≡ , , n, m, n, n , , , , (, , ), , 53., , Circle, , Q1
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Q2, Sol., , Find the pole of the straight line 48x – 54y + 53 = 0 with respect to the circle, 3x2 + 3y2 + 5x – 7y + 2 = 0., 53 , A point on polar is A 0,, , 54 , now chord of contact w.r.t. A, 53 5, 7, 53 , 3x 0 + 3y , x + 0 − y +, +, +2=0, 2, 54 , 54 2, , ( ), , (, , ), , 5, 5, 154, x− y−, =0, 2, 9, 108, 270x – 60y – 154 = 0 ...(1), 53 , another point on polar B −, , 0, 48 , equation of chord of contact, −53 , 5, 53 7, 3x , y+0 +2=0, + 3y 0 + x −, −, 2, 48 2, 48 , , ( ), , (, , ), , 13, 7, 73, x+ y+, = 0 ⇒ 78x + 336y + 73 = 0 ...(2), 16, 2, 96, 1 −1 , On solving (1) and (2), pole: , , 2 3 , , Q3, Sol., , Q4, Sol., , Find the polar of the point (a, –b) with respect to the circle, x2 + y2 + 2ax – 2by + a2 – b2 = 0., Polar is, x(a) + y(–b) + a(x + a) – b(y – b) + a2 – b2 = 0, 2ax – 2by + 2a2 = 0, ⇒ ax – by + a2 = 0, Prove that the polar of a given point with respect to any one of the circles, x2 + y2 –2kx + c2 = 0 where k is variable, always passes through a fixed point,, whatever be the value of k., Let point be A(x1, y1), Equation of polar xx 1 + yy 1 − k x + x 1 + c2 = 0, , (, , xx + yy + c ) − k (x + x ) = 0, (, , , , , , 1, , 1, , Circle, , L1, , 2, , ), , 1, , L2, , which always passes through point of intersection of L1 and L2, 54.
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Conjugate Points, Two points P and Q are said to be conjugate, of each other w.r.t. the circle if polar of ‘P’, passes through ‘Q’ and vice-versa., Conjugate Lines, Two lines 1 and 2 are conjugate of each, other if pole of one lies on other and viceversa., Examples :, , Sol., , Q2, Sol., , Find the value of ‘k’ for which the points (2, k) and (k, 3) are conjugate of each, other w.r.t. the circle x2 + y2 = 10., Let P(2, k) and Q(k, 3), polar of P : 2x + ky = 10, also Q lies on it, ⇒ 2k + 3k = 10 ⇒ k = 2, Prove that if two lines at right angles are conjugate with respect to a circle,, one of them must pass through the centre., Let circle be x2 + y2 = a2, let 1 : mx − y = c1, 2 : x + my = c2, for 1 , let pole P(x1, y1), xx 1 + yy 1 = a2, , mx – y = c1, ma2 −a2 , x1, y, a2, ,, = 1 =, ⇒ P, , c, m −1 c1, c1 , 1, , which lies on 2, −a2 , ma2, + m, = c2 ⇒ c2 = 0, c , c1, 1 , ⇒ 2 passes from centre (0, 0), , 55., , Circle, , Q1
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(3) If 2 circles touches internally, then d = r1 − r2, One common tangent (DCT), , (4) If 2 circles intersect each other, then r1 − r2 < d < r1 + r2, Two common tangents (2 DCT), , (5) If one circle lies completely inside of another, circle then d < r1 − r2, No common tangent., , Examples :, , Sol., , (, , ) + ( y − 3), , Find the range of ‘r’ so that the circles x − 1, , ( x − 4 ) + ( y − 1), 2, , 2, , 2, , 2, , = r2 and, , = 9 intersects at 2 distinct points., , ( ), , C1 1, 3 , r1 = |r|, , C2(4,1), r2 = 3, , d = C1C2 = 9 + 4 =, , 13, , r1 − r2 < d < r1 + r2, r −3 <, , 13 < r + 3, , ⇒ − 13 < r − 3 <, , 13 ⇒ 3 − 13 < r < 3 + 13, 57., , Circle, , Q1
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), , (, , ), , 13 < r + 3 ⇒ r >, , 13 − 3, , (, , , − 3 + 13 < r < 3 + 13 …(1), ⇒, , ) (, , (, , , r ∈ −∞, 3 − 13 ∪, (1) ∩ (2), , ((, , ), , 13 − 3, ∞ …(2), , ), , ) (, , , r ∈ − 3 + 13 , 3 − 13 ∪, , Q2, , ), , 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 its centre is, , {( x, y ) : x, (C) {( x, y ) : x, (A), , Sol., , 13 − 3, 3 + 13, , 2, , 2, , } {( x, y ) : y ≤ 0}, = y} ∪ {( 0, y ) : y ≤ 0}, , = 4y ∪, , (B), (D), , {(, , ), , {( x, y ) : x, , 2, , }, , {( x, y ) : y ≤ 0}, = 4y} ∪ {( 0, y ) : y ≤ 0}, (, , ), , x, y : x2 + y – 1, , 2, , =4 ∪, , (D), C1(0, 1), r1 = 1, Let C2(x, y), r2 = |y|, Now circles touch externally then, C1C2 = r1 + r2, , (, , ), , x2 + y − 1, , 2, , = 1+ y, , on squaring, x2 + y 2 − 2y + 1 = 1 + y 2 + 2 y, x2 − 2y = 2 y, , for y > 0 ⇒ x2 = 4y, for y ≤ 0 ⇒ x2 – 2y = –2y ⇒ x = 0, , Q3, Sol., , (, , ) + ( y − 3), , Find common tangent to the circles x2 + y2 = 1 and x − 1, C1 (0, 0), C2 (1, 3), C1C2 =, , 2, , 2, , = 4., , r1 = 1, r2 = 2, 1+9 =, , 10 > r1 + r2, , circles are separated ⇒ 4 common tangents, , Circle, , TCT :, ∆C1MP ∆C2NP, , 58.
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∴ C1P : C2P = r1 : r2 = 1 : 2, 1 3, 1 , P , ≡ P , 1, 3 3, 3 , Let tangent, , 1, y − 1 = mx − , 3, , , (, , ), , ⇒ mx − y + 1 −, , m, = 0 ...(1), 3, , now, , p = r1, m, 2, 1−, 3 = 1 ⇒ 1 − m = m2 + 1, , , 3, , m2 + 1, , 1+, , m2 2m, 8, −2m, −, = m2 + 1 ⇒ m2 =, 9, 3, 9, 3, , −3, 4, in (1), m = 0 ⇒ y = 1, −3, −3, 1, ⇒, m=, x − y + 1+ = 0, 4, 4, 4, 3x +4y = 5, DCT:, ∆C1MQ ∼ ∆C1NQ, m = 0,m =, , Q divides C1C2 in r1 : r2 externally, Q(–1, –3), Let tangent be, y + 3 = m x + 1 ⇒ mx − y + m − 3 = 0, , (, , ), , (, , ), , , , …(2), , now p = r1, 2, , m +1, , (, , = 1⇒ m−3, , ), , 2, , = m2 + 1, , m2 – 6m + 9 = m2 + 1, 4, ⇒ m = , m = N.D., 3, equation (from 2), 4, 4, 4, for m = ⇒ x − y + − 3 = 0, 3, 3, 3, , 4x – 3y – 5 = 0, for m = N.D. ⇒ x = –1, 59., , Circle, , m−3
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Radical Axis, Radical axis of 2 circles is the locus of a point, whose powers w.r.t. the two circles are equal., The equation of radical axis of two circles, S1 = 0 and S2 = 0 is given by S1 – S2 = 0, , Circle, , Note :, (a) , If two circles intersect, then the radical axis, is the common chord of the two circles., (b) , If two circles touch each other then the, radical axis is the common tangent of the, two circles at the common point of contact., (c) , Radical axis is always perpendicular to the, line joining the centres of the two circles., (d) , Radical axis need not always pass through, the mid point of the line joining the centres, of the two circles., (e) , Radical axis bisects a common tangent, between the two circles., (f) , If one circle is contained in another circle, when radical axis passes outside to both the, circles., (g) , Concentric circles do not have radical axis., , 60.
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Q6, Sol., , Find the equation of the circle which bisects the circumference of the circle, x2 + y2 + 2y – 3 = 0 and touches the line x – y = 0 at origin., Equation of circle is S + λL = 0, , (, , ), , S1 : x 2 + y 2 + λ x − y = 0, , Also given circle S2 : x2 + y2 + 2y – 3 = 0, S1 – S2 = 0 ⇒ λx − λy − 2y + 3 = 0, which is diameter of S2 = 0 hence, λ 0 − λ −1 − 2 −1 + 3 = 0, , ( ), , ( ), , ( ), , λ=–5, Equation : x2 + y2 –5x + 5y = 0, , Radical centre :, Point of intersection of the radical axis of, 3 circles taken 2 at a time is called Radical, Centre., , Note :, (i) Radical axis taken 2 at a time will be concurred, at a point., (ii) Radical centre of three circles described on, sides of a ∆ as diameter is orthocentre of the, ∆., , Coaxial system of circles :, , Definition, , 63., , Circle, , A system of circles, every 2 of, which have the same radical, axis, is called coaxial system of, circles.
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Examples :, , Q1, Sol., , Find the equation of the circle passes through (1, 1) belonging to the system of, coaxial circles which touches x2 + y2 = 8 at (2, 2)., Equation of tangent of given circle at (2, 2) is, 2x + 2y = 8 ⇒ x + y = 4, Now circle touching this line at (2, 2) is, , ( x − 2) + ( y − 2), 2, , 2, , (, , ), , +λ x+y−4 =0, , on satisfying this by point (1, 1), 1 + 1 + λ(–2) = 0 ⇒ λ = 1, , (, , circle : x − 2, , ) + ( y − 2) + ( x + y − 4 ) = 0, 2, , 2, , x2 + y 2 − 3x − 3y + 4 = 0, , Q2, , From a point P tangents drawn to the circles, x2 + y 2 + x − 3 = 0, 3x2 + 3y 2 − 5x + 3y = 0, and 4x2 + 4y 2 + 8x + 7y + 9 = 0, are of equal length. Find the equation of the circle which passes through P, and which touches the line x + y = 5 at (6, –1)., , Sol., , P will be radical centre hence, S1 : x 2 + y 2 + x − 3 = 0, S2 : x2 + y 2 −, , 5, x+y =0, 3, , S3 : x2 + y 2 + 2x +, , 7, 9, y+ =0, 4, 4, , Now radical axis, 8, S1 − S2 = 0 ⇒ x − y − 3 = 0, 3, 8x − 3y = 9 ...(1), S1 − S3 = 0 ⇒ −x −, , 7, 9, y−3− = 0, 4, 4, , Circle, , 4x + 7y + 21 = 0 …(2), , 64.
