P is a point outside a circle and is 13 cm away from its centre. A secant drawn from the point P intersects the circle at points A and B in such a way that PA = 9 cm and AB = 7 cm. It is known that PA < PB. The radius of the circle is :

PC = 13 cm, PA = 9 cm and AB = 7 cm.
From the external point P we have drawn a tangent at point L. Then we have drawn CL.
According to the property of tangent [A tangent to a circle is perpendicular to the radius at the point of tangency.] we can say, PL ⏊ LC.
∴For ΔPLC, PL2 + LC2 = PC2 ......equ(1)
We know that, if a secant segment and tangent segment are drawn to a circle from the same external point, the product of the length of the secant segment and its external part equals the square of the length of the tangent segment.
According to this property: (PL)2 = PA × PB
(PL)2 = PA × (PA + AB)
(PL)2 = 9 × (9 + 7)
(PL)2 = 144
From (1) we can say,
144 + LC2 = 132
LC2 = 169 - 144 = 25
LC = 5 cm.