Question

Let PQ and RSbe 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. $\frac{\sqrt{PQ^{2} + RS^{2}}}{2}$

Answer 1

Answer: (A)

$\sqrt{PQ.RS}$

Answer 2

$\tan\theta = \frac{PQ}{PR} = \frac{PQ}{2r}$

Also $\tan\left( \frac{\pi}{2} - \theta \right) = \frac{RS}{2r}$

i.e. $\cot\theta = \frac{RS}{2r}$

$\therefore$ $\tan\theta.\cot\theta = \frac{PQ.RS}{4r^{2}}$

$\Rightarrow 4r^{2} = PQ.RS \Rightarrow 2r = \sqrt{(PQ)(RS)}$