Question

In the given figure, PT is a tangent to a circle of radius 6 cm. If P is at a distance of 10 cm from the center O and PB = 5 cm, then what is the length of chord BC?

(a) 7.8 cm 

(b) 8 cm 

(c) 8.4 cm 

(d) 9 cm

Answer 1

Answer : (a) 7.8 cm

Given 

PO = 10 cm, radius OT = 6 cm, PB = 5 cm 

In rt. ∆ OTP, (∠OTP = 90° radius OT ⊥ tangent PT)

PT = \(\sqrt{OP^2 -OT^2}\) 

= \(\sqrt{100-36}\) = \(\sqrt{64}\) = 8

 ∴ By Tangent – Secant Theorem 

PT²= PB × PC 

⇒ 8² = 5 × (BC + PB) 

⇒ 64 = 5 (BC + 5) 

⇒ 5BC = 39 

⇒ BC = 7.8 cm.