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

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$\sqrt{P Q \cdot R S}$

$\frac{P Q + R S}{2}$

$\frac{2 P Q \cdot R S}{P Q + R S}$

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

answer is A.

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Detailed Solution

Let $∠ S P R = θ$

Then $∠ Q R P = (\frac{p}{2} - θ), ∠ P Q R = θ$

In $Δ P Q R, \tan (\frac{π}{2} - θ) = \frac{P Q}{P R}$

$\Rightarrow \cot θ = \frac{P Q}{2 r}$ …(i)

Also, $in Δ P R S, \tan θ = \frac{R S}{P R} = \frac{R S}{2 r}$ …(ii)

From Eqs (i) and (ii), we get

$\begin{array}{l} \frac{P Q}{2 r} \cdot \frac{R S}{2 r} = 1 \\ \Rightarrow 4 r^{2} = P Q \cdot R S \\ \Rightarrow 2 r = \sqrt{P Q \cdot R S} \end{array}$

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