# NCERT Solutions for Class 10 Maths Chapter 8 Circles

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.