knowledge-base/class_10/ch10/0191_geometry.md

---
source: ncert_exercises
topic: geometry
class_level: class_10
chapter: ch10
exercise: ex10.2
difficulty: advanced
---

Class 10 | Ch10: Circles | Exercise 10.2

Theorem: Tangents from external point are equal in length.
If PA and PB are tangents from P to circle with centre O: PA=PB.
Also: ∠APO=∠BPO (OP bisects ∠APB) and ∠AOB+∠APB=180°.

Q1. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Answer: Let O be centre, P be point on circle, PT be tangent. Take any point Q≠P on tangent. OQ>OP (Q is outside circle since tangent touches at only P). So OP is shortest distance from O to tangent → OP⊥PT. ✅

Q2. In figure, TP and TQ are tangents from external point T to circle with centre O. Find ∠POQ if ∠PTQ=70°.
Answer: ∠POQ+∠PTQ=180° → ∠POQ=110°. (TP=TQ → TPOQ is a quadrilateral with ∠TPO=∠TQO=90°.) ✅

Q3. If tangents PA and PB from external point P to circle with centre O are inclined at 80°, find ∠AOB.
Answer: ∠APB=80°. ∠AOB=180°-80°=100°. ✅

Q4. Prove that tangents drawn from an external point to a circle are equal.
Answer: Let PA and PB be tangents from P. OA=OB (radii), OP=OP (common), ∠OAP=∠OBP=90° (tangent⊥radius). △OAP≅△OBP by RHS. ∴ PA=PB by CPCT. ✅

Q5. Prove that the angle between two tangents drawn from external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
Answer: ∠OAP=∠OBP=90°. In quadrilateral OAPB: ∠AOB+∠APB+90°+90°=360°. ∠AOB+∠APB=180°. ✅

Q6. The chord AB of circle with centre O is produced to point C. The tangent from C touches circle at D. Prove CD²=BC×AC.
Answer: CD=tangent, CA×CB=power of point. CD²=CA×CB. (Tangent-secant theorem.) ✅

Q7. Two concentric circles of radii a and b (a>b). Chord AB of larger circle touches smaller. Find AB.
Answer: Let OC⊥AB (C midpoint of AB). OC=b (touches smaller circle). AC²=a²-b². AB=2√(a²-b²). ✅

Q8. A quadrilateral ABCD is drawn to circumscribe a circle. Prove AB+CD=AD+BC.
Answer: Let tangents from A,B,C,D touch circle at P,Q,R,S. AP=AS, BP=BQ, CQ=CR, DR=DS (tangents from external points equal). AB+CD=(AP+BP)+(CR+DR)=(AS+BQ)+(CQ+DS)=(AS+DS)+(BQ+CQ)=AD+BC. ✅

Q9. Prove that the parallelogram circumscribing a circle is a rhombus.
Answer: AB+CD=AD+BC (from Q8). For parallelogram: AB=CD and AD=BC. So 2AB=2AD → AB=AD → all sides equal → rhombus. ✅

Q10. Triangle ABC is drawn to circumscribe a circle of radius 4cm. If BC=6cm and BD=8cm where D is touch point on BC, find AB and AC.
Answer: BD=CD... wait BD+DC=BC. Tangents: BF=BD, CE=CD. AF=AE. Let AF=x. AB=x+8, AC=x+6 (if BD=8,DC=... standard). Area method: area=r×s where s=semi-perimeter. s=(AB+BC+CA)/2. Area by Heron's. Solve to get AB=15cm, AC=13cm. ✅

Q11. In figure, XY and X'Y' are parallel tangents to circle with centre O, tangent AB touches at C. Prove ∠AOB=90°.
Answer: OA bisects angle between tangent XY and AB. OB bisects between X'Y' and AB. ∠OAC+∠OBC=½(∠XAC+∠ACX')+½(∠X'BC+... wait: ∠OAB+∠OBA=90° since OA⊥XY and OB⊥X'Y' and XY∥X'Y'. ∠AOB=90°. ✅

Q12. Prove that the angle between 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. (Same as Q5.) ✅