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. $\sqrt{PQ \cdot RS}$
• B. $\frac {PQ+RS}{2}$
• C. $\frac {2PQ \cdot RS}{PQ+RS}$
• D. $\sqrt{\frac {{PQ}^2+{RS}^2}{2}}$

Answer 1

Answer: (A)

From figure, it is clear that $\triangle PRQ$ and $\triangle RSP$ are similar,

$\therefore \frac {PR}{RS}=\frac {PQ}{RP}\Rightarrow {PR}^2=PQ . RS$
$\Rightarrow PR=\sqrt{PQ \cdot RS}$
$\Rightarrow 2r=\sqrt{PQ \cdot RS}$