Question

Let $PQ$ and $RS$ be tangents at the extremities of the diameter $PR$ of a circle of radius $r$. If $PS$ and $RQ$ intersect at a point $X$ on the circumference of the circle, then $2r$ equals

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

Answer 1

The correct option is A $\sqrt{PQ\times RS}$


From the above figure, we have $\frac{PQ}{PR}=\tan (\frac{\pi }{2}-\theta )=\cot \theta$
and $\frac{RS}{PR}=\tan \theta$
$\Rightarrow \frac{PQ}{PR}.\frac{RS}{PR}=1$
$\Rightarrow (PR{)}^2=PQ.RS$
$\Rightarrow (2r{)}^2=PQ.RS$
$\Rightarrow 2r=\sqrt{PQ.RS}$