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.