In the figure (i) given below, chord AB and diameter CD of a circle with centre O meet at P. PT is tangent to the circle at T. If AP = 16 cm, AB = 12 cm and DP = 2 cm, find the length of PT and the radius of the circle.

We know that if a chord and a tangent intersect externally, then the product of the lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

∴ PA.PB = PT² …..(i)

From figure,

PB = PA - AB = 16 - 12 = 4 cm.

Putting values in equation (i),

16 x 4 = PT²
PT² = 64
PT = $\sqrt{64}$ = 8 cm.

Join OT as shown in the figure below:

In △OTP,

OT ⊥ TP (∵ tangents and radius at the point of contact are perpendicular to each other.)

In right angled triangle OTP,

OP² = OT² + PT² (By pythagoras theorem)
(OD + DP)² = OT² + PT²

Since, OD = OT = radius of circle = r.

(r + 2)² = r² + 8²
r² + 4 + 4r = r² + 64
r² - r² + 4r = 64 - 4
4r = 60
r = 15 cm.

Hence, the length of PT = 8 cm and radius of circle = 15 cm.