Q. Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
- Ans. Given, a circle with centre O touches the opposite sides of a quadrilateral at points P, Q, R and S.
- In order to prove the above, the following need to be satisfied :
- ∠AOB + ∠COD = 180° and ∠AOD + ∠BOC = 180°
- Join OS, OP, OQ and OR.
- In DOQC and DOCR,
- QC = CR
- [Tangents from external point C]
- OC = OC [Common]
- OQ = OR [Radii of circle]
- Thus DOQC ≅ DOCR [by S.S.S. congruency]
- Hence ∠QOC = ∠COR
- [Corresponding angles of congruent triangles]
- Similarly,
- ∠ROB = ∠SOB, ∠SOA = ∠POA
- and ∠POD = ∠QOD
- Now ∠QOC + ∠COR + ∠ROB + ∠SOB + ∠SOA
- + ∠POA + ∠POD + ∠QOD = 360°
- ⇒ 2(∠QOC + ∠ROB + ∠SOA + ∠POD) = 360°
- and 2(∠COR + ∠SOB + ∠POA + ∠QOD) = 360°
- ⇒ ∠QOC + ∠ROB + ∠SOA + ∠POD = 180°
- and ∠COR + ∠SOB + ∠POA + ∠QOD = 180°
- ⇒ ∠QOC + ∠SOB + ∠SOA + ∠QOD = 180°
- and ∠COR + ∠ROB + ∠POA + ∠POD = 180°
- ⇒ ∠QOC + ∠QOD + ∠SOA + ∠SOB = 180°
- and ∠COR + ∠ROB + ∠POA + ∠POD = 180°
- ⇒ ∠AOB + ∠COD = 180°
- and ∠AOD + ∠BOC = 180°
- Hence Proved.
Q. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 13 cm. Find the length of PQ.
Ans. 12 cm
Q. PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents P and Q intersect at a point T. Find the length of TP.
Ans. 6.67 cm.
Q. Two concentric circles are of radii 10 cm and 6 cm. Find the length of the chord of the larger circle which touches the smaller circle.
- Ans. 16 cm.