Question

Let $PQ$ and $RS$ be the 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 the value of $2r$ is equal to

• 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. $\sqrt{\frac{P Q^{2} + R S^{2}}{2}}$

Answer 1

Answer: (A)

From the figure, it is clear that $\Delta PRQ$ and $\Delta RSP$ are similar.

$\therefore tan\theta =\frac{P R}{R S}=\frac{P Q}{R P}$
$\Rightarrow PR^{2}=PQ\cdot RS$
$\Rightarrow PR=\sqrt{P Q \cdot R S}$
$\Rightarrow 2r=\sqrt{P Q \cdot R S}$