Class 10 Maths Chapter – 10 Circle EXERCISE 10.2 Questions and Answers
EXERCISE 10.2
Q1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is:
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
Answer: (A) 7 cm.
Given: Tangent length TQ = 24 cm, OQ = 25 cm, T is the point of contact.
Proof: By Theorem 10.1, OT ⊥ TQ, so triangle OTQ is right-angled at T:
Hence r=7cm.
Q2. If TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to:
(A) 60°
(B) 70°
(C) 80°
(D) 90°
Answer: (B) 70°.
Given: Tangents TP, TQ from external point T; ∠POQ = 110°.
Proof: By Theorem 10.1, ∠OPT = ∠OQT = 90°.
In quadrilateral OPTQ, angles sum to 360°:
∠PTQ=360°−(90°+90°+110°)=70°
Q3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to:
(A) 50°
(B) 60°
(C) 70°
(D) 80°
Answer: (A) 50°.
Given: Tangents PA, PB with ∠APB = 80°.
Proof: ∠OAP = ∠OBP = 90°,
so in quadrilateral OAPB:
∠AOB=360°−(90°+90°+80°)=100°
By the remark on Theorem 10.2, OP bisects ∠AOB,
So: ∠POA=100°/2=50°
Q4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Answer: Given: A circle with centre O and diameter AB; tangents drawn at A and at B.
To Prove: The two tangents are parallel.
Proof: By Theorem 10.1, the radius through the point of contact is perpendicular to the tangent.
So the tangent at A is perpendicular to AB, and the tangent at B is also perpendicular to AB.
Since both tangents are perpendicular to the same line AB, they are parallel to each other. Hence proved.
Q5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Answer: Given: A circle with centre O, tangent XY touching the circle at point P.
To Prove: The perpendicular to XY at P passes through O.
Proof: By Theorem 10.1, OP ⊥ XY, i.e., OP is perpendicular to the tangent at the point of contact. Through a given point on a line, only one perpendicular can be drawn.
Therefore the perpendicular to the tangent at P must be the line OP, which passes through the centre O. Hence proved.
Q6. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Answer: Given: OA = 5 cm, tangent length AT = 4 cm, T is the point of contact.
Proof: ∠OTA = 90° (radius ⊥ tangent).
Therefore:
So the radius = 3 cm.
Q7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Answer: Given: Concentric circles with common centre O; radii 5 cm and 3 cm; chord AB of the larger circle touches the smaller circle at P.
To Find: AB.
Construction: Join OP and OA.
Proof: Since AB is a tangent to the smaller circle at P, OP ⊥ AB (Theorem 10.1) and OP = 3 cm. Also OA = 5 cm.
The perpendicular from the centre bisects the chord, so AP = PB.
So AB=2×4= 8 cm.
Q8. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.
Answer: Given: A quadrilateral ABCD circumscribing a circle with centre O.
To Prove: AB + CD = AD + BC.
Construction: Let the circle touch sides AB, BC, CD, DA at P, Q, R, S respectively. Proof: By Theorem 10.2, tangents from an external point are equal:
- From A: AP = AS
- From B: BP = BQ
- From C: CQ = CR
- From D: DR = DS
Adding all four:
(AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)
That is, AB + CD = AD + BC. Hence proved.
Q9. XY and X′Y′ are two parallel tangents to a circle with centre O, and another tangent AB with point of contact C intersects XY at A and X′Y′ at B. Prove that ∠AOB = 90°.
Answer: Given: Two parallel tangents XY and X′Y′ to a circle with centre O; a third tangent AB touching the circle at C, meeting XY at A and X′Y′ at B. To Prove: ∠AOB = 90°.
Construction: Join OC, and join O to the points of contact P and Q of the parallel tangents XY and X′Y′.
Proof: Since XY ∥ X′Y′, the radii OP and OQ to their points of contact lie on the same straight line, so ∠POQ = 180°. Now OP ⊥ AC and OQ ⊥ BC (radius ⊥ tangent).
In triangles AOP and AOC: AP = AC (tangents from A), OP = OC (radii), AO is common. So ∆AOP ≅ ∆AOC (SSS), giving ∠AOP = ∠AOC. Similarly, ∆BOC ≅ ∆BOQ, giving ∠BOC = ∠BOQ.
Therefore:
Hence proved.
Q10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
Answer: Given: Tangents PA and PB from an external point P, touching the circle with centre O at A and B.
To Prove: ∠APB + ∠AOB = 180°.
Construction: Join OA, OB, OP. Proof: ∠OAP = ∠OBP = 90° (radius ⊥ tangent, Theorem 10.1).
In quadrilateral OAPB, the sum of angles is 360°:
∠APB+∠AOB=360°−(90°+90°)=180°
So ∠APB and ∠AOB are supplementary. Hence proved.
Q11. Prove that a parallelogram circumscribing a circle is a rhombus.
Answer: Given: A parallelogram ABCD circumscribing a circle.
To Prove: ABCD is a rhombus.
Construction: Let the circle touch sides AB, BC, CD, DA at P, Q, R, S respectively.
Proof: By Theorem 10.2: AP = AS, BP = BQ, CR = CQ, DR = DS. Adding: AB + CD = AD + BC (as in Q8). But ABCD is a parallelogram, so AB = CD and AD = BC.
Therefore: 2AB=2AD ⟹ AB=AD
So all four sides are equal (opposite sides of a parallelogram are equal). Hence ABCD is a rhombus. Hence proved.
Q12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.
Answer: Given: Triangle ABC circumscribing a circle of radius 4 cm; the circle touches BC at D with BD = 8 cm and DC = 6 cm.
To Find: AB and AC.
Construction: Let the circle touch AB at F and AC at E. Join the centre to the points of contact.
Proof: Tangents from the same point are equal (Theorem 10.2):
- From B: BD = BF = 8 cm
- From C: CD = CE = 6 cm
- From A: AF = AE = x (say)
So AB = x + 8, AC = x + 6, BC = 14 cm. Semi-perimeter:
Area of triangle = r × s = 4(x + 14). By Heron’s formula:
Equating the two:
Since x + 14 ≠ 0: 16(x + 14) = 48x, so x + 14 = 3x, giving x = 7.
Therefore AB = 7 + 8 = 15 cm and AC = 7 + 6 = 13 cm.
Q13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Given: A quadrilateral ABCD circumscribing a circle with centre O.
To Prove: ∠A + ∠D = 180° and ∠B + ∠C = 180° (i.e., opposite sides subtend supplementary angles at the centre).
Construction: Let the circle touch AB, BC, CD, DA at P, Q, R, S respectively. Join OP, OQ, OR, OS.
Proof: Each radius is perpendicular to the side it meets (Theorem 10.1). In quadrilateral APOS: ∠APO = ∠ASO = 90°,
So ∠A+∠POS=180°⟹∠POS=180°−∠A
Similarly: ∠POQ = 180° − ∠B; ∠QOR = 180° − ∠C; ∠ROS = 180° − ∠D. Now ∠POS + ∠ROS = ∠POR and ∠POQ + ∠QOR = ∠POR (both pairs make the same angle around O through P and R). So:
(180°−∠A)+(180°−∠D)=(180°−∠B)+(180°−∠C)
360°−(∠A+∠D)=360°−(∠B+∠C)
∠A+∠D=∠B+∠C
Since all four angles total 360°, each pair sums to 180°:
∠A+∠D=180°and∠B+∠C=180°
Hence opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre. Hence proved.