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.
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  • 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.