Question

$A$ point $P$ is outside a circle and is 13 inches from the centre. A secant from $P$ cuts the circle at $Q$ and $R$ so that the external segment of the secant $PQ$ is 9 inches and $QR$ is $7$ inches. The radius of the circle is :

• A. $3$ inches
• B. $4$ inches
• C. $5$ inches
• D. $6$ inches
• E. $7$ inches

Answer 1

The correct option is C $5$ inches

Let $\angle APR=\alpha$

Now consider triangle $APR$ :

$\Rightarrow \cos \alpha =\frac{{13}^2+{16}^2-r^2}{2\times 13\times 16}$ $\to$equation (1)

Consider triangle $APQ$ :

$\Rightarrow \cos \alpha =\frac{{13}^2+9^2-r^2}{2\times 13\times 9}$ $\to$equation (2)

Equating equation (1) and (2);

$\Rightarrow \frac{{13}^2+{16}^2-r^2}{2\times 13\times 16}=\frac{{13}^2+9^2-r^2}{2\times 13\times 9}$

$\Rightarrow r^2={13}^2-144$

$\Rightarrow r^2=25$

$\therefore r=5$