In the given figure, TB is a chord which passes through the centre of the circle. PT is a tangent to the circle at the point T on the circle. If PT = 10 cm, PA = 5 cm and AB = x cm, then the radius of the circle is:

$5\sqrt{3}$ cm

We know that,

(Hypotnuese)² = (Perpendicular)² + (Base)²

We have,

PT = 10 cm 

PA = 5 cm 

We also know that,PT² = AP × PB

 PB = PA + AB = 5 + x 

100 = 5 × (5 + x) 

x = 15 

PB = 20 

So,  ∠PTB = 90° 

20² = 10² + BT²

BT = 10√3 

Radius of circle = 10√3/2 =  5√3