In the given figure, AB is diameter and PC is tangent to the circle with center O. If AP = 40 cm, CP = 20 cm, find the radius of the circle.

Let radius of circle be r cm.

From figure,

⇒ AP = AB + BP

⇒ 40 = AO + OB + BP

⇒ 40 = r + r + BP

⇒ 40 = 2r + BP

⇒ BP = (40 - 2r) cm.

We know that,

Tangent at point of intersection and radius of circle are perpendicular to each other.

In △OCP,

⇒ OP² = OC² + PC²

⇒ (OB + BP)² = OC² + PC²

⇒ (r + 40 - 2r)² = r² + 20²

⇒ (40 - r)² = r² + 400

⇒ 1600 + r² - 80r = r² + 400

⇒ r² - r² + 80r = 1600 - 400

⇒ 80r = 1200

⇒ r = $\dfrac{1200}{80}$ = 15 cm.

Hence, radius of circle = 15 cm.