In a circle with centre O, PA and PB are tangents drawn from an external point P. If PA = 9 cm and ∠APB = 60°, find the :

(a) We know that,

The lengths of two tangents drawn from an external point to a circle are equal.

∴ PB = PA

⇒ PB = 9 cm.

Hence, length of PB = 9 cm.

(b) Join OA, OB and OP.

Figure unavailable in the source: In a circle with centre O, PA and PB are tangents drawn from an external point P. If PA = 9 cm and ∠APB = 60°, find the:. Specimen 2027, ICSE Specimen 2027 Maths Solved Question Paper.

We know that,

The tangents drawn from an external point are equally inclined to the line joining the centre of the circle and that point.

∴ OP bisects ∠APB.

⇒ ∠OPA = $\frac{1}{2} \times ∠APB = \frac{1}{2} \times 60° = 30°$

We know that,

Radius and tangent at the point of contact are perpendicular to each other.

∴ ∠OAP = 90°

In right angled triangle OAP,

$$\begin{aligned}\Rightarrow \tan 30° = \frac{OA}{PA} \\[1em] \Rightarrow OA = PA \times \tan 30° \\[1em] \Rightarrow OA = 9 \times \frac{1}{\sqrt{3}} \\[1em] \Rightarrow OA = \frac{9 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\[1em] \Rightarrow OA = \frac{9\sqrt{3}}{3} \\[1em] \Rightarrow OA = 3\sqrt{3} \\[1em] \Rightarrow OA = 3 \times 1.7321 \\[1em] \Rightarrow OA = 5.1963 \\[1em] \Rightarrow OA = 5.2 \text{ cm}.\end{aligned}$$

Hence, radius of the circle = 5.2 cm.