Q. Let $PQ$ and $PS$ 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) $\sqrt{PQ.RS}$
B) $\frac{PQ + RS}{2}$
C) $\frac{2PQ.RS}{PQ + RS}$
D) $\sqrt{\frac{PQ^{2} + RS^{2}}{2}}$
Explanation
(a)
Let $RS$ and $PQ$ are the tangents at the extremities of diameter of circle
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$In\ \mathrm{\Delta}\ RQP,\ \tan\theta = \frac{PQ}{PR} = \frac{PQ}{2r}\ \ \ \ \ldots(i)$
Also, in $\mathrm{\Delta}SRP$,
$\tan\left( \frac{\pi}{2} - \theta \right) = \frac{RS}{RP} = \frac{RS}{2r}$
$\Rightarrow \cot\theta = \frac{RS}{2r}\ \ \ \ldots(ii)$
From Eqs. (i) and (ii), we get
$\tan\theta.\cot\theta = \frac{PQ.RS}{4r^{2}}$
$\Rightarrow 4r = PQ.RS$
$\Rightarrow 2r = \sqrt{(PQ)(RS)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.