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NCERT Exemplar Solutions For Class 10 Maths Chapter 9Circles, , Exercise 9.1, , Page No: 102, , Choose the correct answer from the given four options in the following questions:, 1. If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle, which is tangent to the other circle is, (A) 3 cm, (B) 6 cm, (C) 9 cm, (D) 1 cm, Solution:, , According to the question,, OA = 4cm, OB = 5cm, And, OA ⊥ BC, Therefore, OB2 = OA2 + AB2, ⇒, 52 = 42 + AB2, ⇒, AB = √(25 – 16) = 3cm, ⇒, BC = 2AB = 2 × 3cm = 6cm, 2. In Fig. 9.3, if AOB = 125°, then COD is equal to, (A) 62.5°, (B) 45°, (C) 35°, (D) 55°

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NCERT Exemplar Solutions For Class 10 Maths Chapter 9Circles, Solution:, ABCD is a quadrilateral circumscribing the circle, We know that, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary, angles at the center of the circle., So, we have, ∠AOB + ∠COD = 180°, , 125° + ∠COD = 180°, ∠COD = 55°, 3. In Fig. 9.4, AB is a chord of the circle and AOC is its diameter such that ACB = 50°. If AT is the, tangent to the circle at the point A, then BAT is equal to, (A) 65°, (B) 60°, (C) 50°, (D) 40°, , Solution:, According to the question,, A circle with centre O, diameter AC and ∠ACB = 50°, AT is a tangent to the circle at point A, Since, angle in a semicircle is a right angle, ∠CBA = 90°, By angle sum property of a triangle,, ∠ACB + ∠CAB + ∠CBA = 180°, 50° + ∠CAB + 90° = 180°, ∠CAB = 40° … (1), Since tangent to at any point on the circle is perpendicular to the radius through point of, contact,, We get,, OA ⏊ AT, ∠OAT = 90°, ∠OAT + ∠BAT = 90°, ∠CAT + ∠BAT = 90°, 40° + ∠BAT = 90° [from equation (1)], ∠BAT = 50°, 4. From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the, pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is

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NCERT Exemplar Solutions For Class 10 Maths Chapter 9Circles, (A) 60 cm2, (C) 30 cm2, , (B) 65 cm2, (D) 32.5 cm2, , Solution:, Construction: Draw a circle of radius 5 cm with center O., Let P be a point at a distance of 13 cm from O., , Draw a pair of tangents, PQ and PR., OQ = OR = radius = 5cm …equation (1), And OP = 13 cm, , We know that, tangent to at any point on the circle is perpendicular to the radius through, point of contact,, Hence, we get,, OQ ⏊ PQ and OR ⏊PR, △POQ and △POR are right-angled triangles., Using Pythagoras Theorem in △PQO,, (Base)2+ (Perpendicular)2= (Hypotenuse)2, (PQ)2 + (OQ)2= (OP)2, (PQ)2 + (5)2 = (13)2, (PQ)2 + 25 = 169, (PQ)2 = 144, PQ = 12 cm, Tangents through an external point to a circle are equal., So,, PQ = PR = 12 cm … (2), Therefore, Area of quadrilateral PQRS, A = area of △POQ + area of △POR, Area of right angled triangle = ½ x base x perpendicular, A = (½ x OQ x PQ) + (½ x OR x PR), A = (½ x 5 x 12) + (½ x 5 x 12), A = 30 + 30 = 60 cm2, , 5. At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle., The length of the chord CD parallel to XY and at a distance 8 cm from A is

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NCERT Exemplar Solutions For Class 10 Maths Chapter 9Circles, (A) 4 cm, (C) 6 cm, , (B) 5 cm, (D) 8 cm, , Solution:, , According to the question,, Radius of circle, AO=OC = 5cm, AM=8CM, AM=OM+AO, OM =AM-AO, Substituting these values in the equation,, OM= (8-5) =3CM, OM is perpendicular to the chord CD., In ∆OCM <OMC=90°, By Pythagoras theorem,, OC²=OM²+MC², Therefore,, CD= 2 ×CM = 8 cm

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NCERT Exemplar Solutions For Class 10 Maths Chapter 9Circles, , Exercise 9.2, , Page No: 105, , Write ‘True’ or ‘False’ and justify your answer in each of the following:, 1. If a chord AB subtends an angle of 60° at the centre of a circle, then angle between the tangents, at A and B is also 60°., Solution:, False, Justification:, For example,, Consider the given figure. In which we have a circle with centre O and AB a chord with, ∠AOB = 60°, , Since, tangent to any point on the circle is perpendicular to the radius through point of, contact,, We get,, OA ⏊ AC and OB ⏊ CB, ∠OBC = ∠OAC = 90° … eq(1), Using angle sum property of quadrilateral in Quadrilateral AOBC,, We get,, ∠OBC + ∠OAC + ∠AOB + ∠ACB = 360°, 90° + 90° + 60° + ∠ACB = 360°, ∠ACB = 120°, Hence, the angle between two tangents is 120°., Therefore, we can conclude that,, the given statement is false., 2. The length of tangent from an external point on a circle is always greater than the radius of the, circle., Solution:

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NCERT Exemplar Solutions For Class 10 Maths Chapter 9Circles, False, Justification:, Length of tangent from an external point P on a circle may or may not be greater than the, radius of the circle., 3. The length of tangent from an external point P on a circle with centre O is always less than OP., Solution:, True, Justification:, Consider the figure of a circle with centre O., Let PT be a tangent drawn from external point P., Now, Joint OT., OT ⏊ PT, , We know that,, Tangent at any point on the circle is perpendicular to the radius through point of contact, Hence, OPT is a right-angled triangle formed., We also know that,, In a right angled triangle, hypotenuse is always greater than any of the two sides of the, triangle., Hence,, OP > PT or PT < OP, Hence, length of tangent from an external point P on a circle with center O is always less, than OP., 4. The angle between two tangents to a circle may be 0°., Solution:, True, Justification:, The angle between two tangents to a circle may be 0°only when both tangent lines, coincide or are parallel to each other.

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NCERT Exemplar Solutions For Class 10 Maths Chapter 9Circles, , 5. If angle between two tangents drawn from a point P to a circle of radius a and centre O is 90°,, then OP = a√2., Solution:, Tangent is always perpendicular to the radius at the point of contact., Hence, ∠OAP = 90, If 2 tangents are drawn from an external point, then they are equally inclined to the line, segment joining the centre to that point., Consider the following figure,, , Therefore, ∠OPA = 12∠APB = 12×60° = 30°, Using angle sum property of triangle in ∆AOP,, ∠AOP + ∠OAP + ∠OPA =180°, ∠AOP + 90° + 30° = 180°, ∠AOP = 60°, So, in △AOP, tan (∠AOP) = AP/ OA, √ 3= AP/a, Therefore, AP = √ 3a, Hence, proved

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NCERT Exemplar Solutions For Class 10 Maths Chapter 9Circles, , Exercise 9.3, , Page No: 107, , 1. Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord AC of, length 8 cm is a tangent to the inner circle. Find the radius of the inner circle., Solution:, , From the figure,, Chord AB = 8 cm, OC is perpendicular to the chord AB, AC = CB = 4 cm, In right triangle OCA, OC2 + CA2 = OA2, OC2 = 52 - 42 = 25 - 16 = 9, OC = 3 cm, 2. Two tangents PQ and PR are drawn from an external point to a circle with centre O. Prove that, QORP is a cyclic quadrilateral., Solution:

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NCERT Exemplar Solutions For Class 10 Maths Chapter 9Circles, 4. Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the, lines., Solution:, , Let the lines be l1 and l2., Assume that O touches l₁ and l₂ at M and N,, We get,, OM = ON (Radius of the circle), Therefore,, From the centre ''O'' of the circle, it has equal distance from l₁ & l₂., In Δ OPM & OPN,, OM = ON (Radius of the circle), ∠OMP = ∠ONP (As, Radius is perpendicular to its tangent), OP = OP (Common sides), Therefore,, Δ OPM = ΔOPN (SSS congruence rule), By C.P.C.T,, ∠MPO = ∠NPO, So, l bisects ∠MPN., Therefore, O lies on the bisector of the angle between l₁ & l₂ ., Hence, we prove that the centre of a circle touching two intersecting lines lies on the, angle bisector of the lines., 5. In Fig. 9.13, AB and CD are common tangents to two circles of unequal radii. Prove that AB =, CD.

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NCERT Exemplar Solutions For Class 10 Maths Chapter 9Circles, , Solution:, According to the question,, AB = CD, , Construction: Produce AB and CD, to intersect at P., Proof:, Consider the circle with greater radius., Tangents drawn from an external point to a circle are equal, AP = CP …(1), Also,, Consider the circle with smaller radius., Tangents drawn from an external point to a circle are equal, BP = BD …(2), Subtract Equation (2) from (1). We Get, AP - BP = CP – BD, AB = CD, Hence Proved.

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NCERT Exemplar Solutions For Class 10 Maths Chapter 9Circles, , Exercise 9.4, , Page No: 110, , 1. If a hexagon ABCDEF circumscribe a circle, prove that AB + CD + EF = BC + DE + FA., Solution:, , According to the question,, A Hexagon ABCDEF circumscribe a circle., To prove:, AB + CD + EF = BC + DE + FA, Proof:, Tangents drawn from an external point to a circle are equal., Hence, we have, AM = RA … eq 1 [tangents from point A], BM = BN … eq 2 [tangents from point B], CO = NC … eq 3 [tangents from point C], OD = DP … eq 4 [tangents from point D], EQ = PE … eq 5 [tangents from point E], QF = FR … eq 6 [tangents from point F], [eq 1]+[eq 2]+[eq 3]+[eq 4]+[eq 5]+[eq 6], AM + BM + CO + OD + EQ + QF = RA + BN + NC + DP + PE + FR, On rearranging, we get,, (AM + BM) + (CO + OD) + (EQ + QF) = (BN + NC) + (DP + PE) + (FR + RA), AB + CD + EF = BC + DE + FA, Hence Proved!, , 2. Let s denote the semi-perimeter of a triangle ABC in which BC = a, CA = b, AB = c. If a circle, touches the sides BC, CA, AB at D, E, F, respectively, prove that BD = s – b., Solution:

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NCERT Exemplar Solutions For Class 10 Maths Chapter 9Circles, , According to the question,, A triangle ABC with BC = a , CA = b and AB = c . Also, a circle is inscribed which, touches the sides BC, CA and AB at D, E and F respectively and s is semi- perimeter of, the triangle, To Prove: BD = s – b, Proof:, According to the question,, We have,, Semi Perimeter = s, Perimeter = 2s, 2s = AB + BC + AC [1], As we know,, Tangents drawn from an external point to a circle are equal, So we have, AF = AE [2] [Tangents from point A], BF = BD [3] [Tangents From point B], CD = CE [4] [Tangents From point C], Adding [2] [3] and [4], AF + BF + CD = AE + BD + CE, AB + CD = AC + BD, Adding BD both side, AB + CD + BD = AC + BD + BD, AB + BC - AC = 2BD, AB + BC + AC - AC - AC = 2BD, 2s - 2AC = 2BD [From 1], 2BD = 2s - 2b [as AC = b], BD = s – b, Hence Proved.

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NCERT Exemplar Solutions For Class 10 Maths Chapter 9Circles, 3. From an external point P, two tangents, PA and PB are drawn to a circle with centre O. At one, point E on the circle tangent is drawn which intersects PA and PB at C and D, respectively. If PA, = 10 cm, find the perimeter of the triangle PCD., Solution:, , According to the question,, From an external point P, two tangents, PA and PB are drawn to a circle with center O., At a point E on the circle tangent is drawn which intersects PA and PB at C and D,, respectively. And PA = 10 cm, To Find : Perimeter of △PCD, As we know that, Tangents drawn from an external point to a circle are equal., So we have, AC = CE [1] [Tangents from point C], ED = DB [2] [Tangents from point D], Now Perimeter of Triangle PCD, = PC + CD + DP, = PC + CE + ED + DP, = PC + AC + DB + DP [From 1 and 2], = PA + PB, Now,, PA = PB = 10 cm as tangents drawn from an external point to a circle are equal, So we have, Perimeter = PA + PB = 10 + 10 = 20 cm, 4. If AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as, shown in Fig. 9.17. Prove that ∠BAT = ∠ACB

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NCERT Exemplar Solutions For Class 10 Maths Chapter 9Circles, , Solution:, According to the question,, A circle with center O and AC as a diameter and AB and BC as two chords also AT is a tangent, at point A, , To Prove : ∠BAT = ∠ACB, Proof :, ∠ABC = 90° [Angle in a semicircle is a right angle], In △ABC By angle sum property of triangle, ∠ABC + ∠ BAC + ∠ACB = 180 °, ∠ACB + 90° = 180° - ∠BAC, ∠ACB = 90 - ∠BAC [1], Now,, OA ⏊ AT [Tangent at a point on the circle is perpendicular to the radius through point of, contact ], ∠OAT = ∠CAT = 90°, ∠BAC + ∠BAT = 90°, ∠BAT = 90° - ∠BAC [2], From [1] and [2], ∠BAT = ∠ACB [Proved], 5. Two circles with centres O and O' of radii 3 cm and 4 cm, respectively intersect at two points P, and Q such that OP and O'P are tangents to the two circles. Find the length of the common chord, PQ., Solution:

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NCERT Exemplar Solutions For Class 10 Maths Chapter 9Circles, , According to the question,, Two circles with centers O and O’ of radii 3 cm and 4 cm, respectively intersect at two, points P and Q, such that OP and O’P are tangents to the two circles and PQ is a common, chord., To Find: Length of common chord PQ, ∠OPO' = 90° [Tangent at a point on the circle is perpendicular to the radius through point, of contact], So OPO is a right-angled triangle at P, Using Pythagoras in △ OPO', we have, (OO')2= (O'P)2+ (OP)2, (OO')2 = (4)2 + (3)2, (OO')2 = 25, OO' = 5 cm, Let ON = x cm and NO' = 5 - x cm, In right angled triangle ONP, (ON)2+ (PN)2= (OP)2, x2 + (PN)2 = (3)2, (PN)2= 9 - x2 [1], In right angled triangle O'NP, (O'N)2 + (PN)2= (O'P)2, (5 - x)2 + (PN)2 = (4)2, 25 - 10x + x2 + (PN)2 = 16, (PN)2 = -x2+ 10x - 9[2], From [1] and [2], 9 - x2 = -x2 + 10x – 9, 10x = 18, x = 1.8, From (1) we have, (PN)2 = 9 - (1.8)2

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NCERT Exemplar Solutions For Class 10 Maths Chapter 9Circles, =9 - 3.24 = 5.76, PN = 2.4 cm, PQ = 2PN = 2(2.4) = 4.8 cm, 6. In a right triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting, the hypotenuse AC and P. Prove that the tangent to the circle at P bisects BC., Solution:, , According to the question,, In a right angle ΔABC is which ∠B = 90°, a circle is drawn with AB as diameter, intersecting the hypotenuse AC at P. Also PQ is a tangent at P, To Prove: PQ bisects BC i.e. BQ = QC, Proof:, ∠APB = 90° [Angle in a semicircle is a right-angle], ∠BPC = 90° [Linear Pair], ∠3 + ∠4 = 90 [1], Now, ∠ABC = 90°, So in △ABC, ∠ABC + ∠BAC + ∠ACB = 180°, 90 + ∠1 + ∠5 = 180, ∠1 + ∠5 = 90 [2], Now,, ∠ 1 = ∠ 3[angle between tangent and the chord equals angle made by the chord in, alternate segment], Using this in [2] we have, ∠3 + ∠5 = 90 [3], From [1] and [3] we have, ∠3 + ∠4 = ∠3 + ∠5, ∠4 = ∠5, QC = PQ [Sides opposite to equal angles are equal]

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NCERT Exemplar Solutions For Class 10 Maths Chapter 9Circles, But Also PQ = BQ [Tangents drawn from an external point to a circle are equal], So, BQ = QC, i.e. PQ bisects BC ., 7. In Fig. 9.18, tangents PQ and PR are drawn to a circle such that ∠RPQ = 30°. A chord RS is, drawn parallel to the tangent PQ. Find the ∠RQS., [Hint: Draw a line through Q and perpendicular to QP.], , Solution:, , According to the question,, Tangents PQ and PR are drawn to a circle such that ∠RPQ = 30°. A chord RS is drawn, parallel to the tangent PQ., To Find : ∠RQS