In the figure (ii) given below, chord AB and diameter CD of a circle meet at P. If AB = 8 cm, BP = 6 cm and PD = 4 cm, find the radius of the circle. Also find the length of the tangent drawn from P to 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,

PA = AB + PB = 8 + 6 = 14 cm.

Putting values in equation (i),

14 x 6 = PT²
PT² = 84
PT = $\sqrt{84} = 2\sqrt{21}$ cm.

Joining OT as shown in the figure below:

In △OTP,

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

In right angled triangle OTP,

$OP^2 = OT^2 + PT^2 (\text{By pythagoras theorem}) \\[1em] (OD + DP)^2 = OT^2 + PT^2 \\[1em]$

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

$\Rightarrow (r + 4)^2 = r^2 + (\sqrt{84})^2 \\[1em] \Rightarrow r^2 + 16 + 8r = r^2 + 84 \\[1em] \Rightarrow r^2 - r^2 + 8r = 84 - 16 \\[1em] \Rightarrow 8r = 68 \\[1em] \Rightarrow r = \dfrac{68}{8} \\[1em] \Rightarrow r = 8.5 \text{ cm.}$

Hence, the radius of the circle = 8.5 cm and length of tangent = $2\sqrt{21}$ cm.