P is a point outside a circle with centre O, and it is 14 cm away from the centre. A secant PAB drawn from P intersects the circle at the points A and B such that PA = 10 cm and PB = 16 cm. The diameter of the Circle is:

12 cm

Given OP = 14 cm, PA = 10 cm, PB = 16 cm

Let radius of circle be r cm.

= OQ = OE = r cm

= PR = (14 - r) and PQ = (14 + r)

As we know,

PE x PQ = PA x PB

= (14 - r)(14 + r)= 10 x 16

= \( { 14}^{2 } \) - \( { r}^{2 } \) = 160

= 196 - \( { r}^{2 } \) = 160

= \( { r}^{2 } \) = 196 - 160 = 36

= r = \(\sqrt {36 }\) = 6 cm

Therefore, Diameter 2 x 6 = 12 cm.