Question

Let $P Q$ and $R S$ be tangents at the extremities of a diameter $P R$ of a circle of radius $r$ such that $P S$ and $R Q$ intersect at a point $X$ on the circumference of the circle, then $2 r$ equals

• A. $\sqrt{P Q \cdot R S}$
• B. $\frac{P Q+R S}{2}$
• C. $\frac{2 P Q \cdot R S}{P Q+R S}$
• D. $\frac{\sqrt{(P Q)^{2}+(R S)^{2}}}{2}$

Answer 1

Answer: (A)

According to the question, from the diagram, in $\triangle PQR$
$\tan \theta=\frac{P Q}{P R}$
$\Rightarrow P R=P Q \cot \theta \ldots$ (i)

and in $\triangle P R S$,
$\tan \left(90^{\circ}-\theta\right)=\frac{R S}{P R}$
$\Rightarrow P R=R S \tan \theta$
$\therefore P Q \cot \theta=R S \tan \theta$
$\Rightarrow \tan \theta=\sqrt{\frac{P Q}{R S}} ... (iii)$
From Eqs. (ii) and (iii), we have
$P R=R S \sqrt{\frac{P Q}{R S}}=\sqrt{P Q \cdot R S}$
$\Rightarrow 2 r=\sqrt{P Q \cdot R S} \because P R=2 r$