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 a point x on the circumference of the circle, then 2r equals-
A) $\sqrt {PQ.RS}$
B) $\dfrac{{PQ + RS}}{2}$
C) $\dfrac{{2PQ + RS}}{{PQ + RS}}$
D) $\sqrt {\dfrac{{P{Q^2} + R{S^2}}}{2}}$

Answer 1

Hint- We will draw the figure of a circle in order to get a better understanding of the requirement of the question. Since r is the radius of the circle, 2r will be the diameter of the circle. Sum of all the angles of the triangle is 180 degrees.

Complete step-by-step answer:

The angle made on point x will be 90 degrees.
Now, let the angle on point R be $\theta$ and the angle on point P be $90 - \theta$.
In $\Delta PRQ$,
$\to \tan \theta = \dfrac{{PQ}}{{PR}}$ (since we know that $\tan \theta = \dfrac{{perpendicular}}{{base}}$)
$\to PR = \dfrac{{PQ}}{{\tan \theta }} \\\ \\\ \Rightarrow PR = PQ\cot \theta \\\$
Let the above equation be equation 1-
$\to PR = PQ\cot \theta$ (equation 1)
In $\Delta SPR$,
$\to \tan \left( {90 - \theta } \right) = \dfrac{{RS}}{{PR}} \\\ \\\ \to PR = \dfrac{{RS}}{{\cot \theta }} \\\ \\\ \Rightarrow PR = RS\tan \theta \\\$ ($\tan \left( {90 - \theta } \right) = \cot \theta$)
Let the above equation be equation 2-
$\to PR = RS\tan \theta$ (equation 2)
From equation 1 and 2 we get-
$\to PR = PR \\\ \\\ \to PQ\cot \theta = RS\tan \theta \\\ \\\ \to \dfrac{{PQ}}{{RS}} = \dfrac{{\tan \theta }}{{\cot \theta }} \\\$
$\to \dfrac{{PQ}}{{RS}} = {\tan ^2}\theta$ ($\dfrac{1}{{\cot \theta }} = \tan \theta$)
$\Rightarrow \tan \theta = \sqrt {\dfrac{{PQ}}{{RS}}}$
Putting this value of $\tan \theta$ in equation 2, we get-
$\to PR = RS\tan \theta \\\ \\\ \to PR = RS\sqrt {\dfrac{{PQ}}{{RS}}} \\\$
$\to PR = \sqrt {RS} .\sqrt {RS} \dfrac{{\sqrt {PQ} }}{{\sqrt {RS} }}$ (Since $RS = \sqrt {RS} .\sqrt {RS}$)
$\Rightarrow PR = \sqrt {RS \times PQ}$
Since, PR is the diameter of the circle and the diameter of the circle is 2r (radius is r), we get-
$\to 2r = PR \\\ \\\ \Rightarrow 2r = \sqrt {RS \times PQ} \\\$
Hence, A is the correct option.

Note: Remember the basic formula of $\tan \theta = \dfrac{{perpendicular}}{{base}}$. It is the formula which is used the most in the above question. Always make a diagram in the starting of the answer of such questions to make them easy to solve and understand. The diagram is necessary.