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) ${\displaystyle \frac{PQ+RS}{2}}$
C) ${\displaystyle \frac{2PQ+RS}{PQ+RS}}$
D) $\sqrt{{\displaystyle \frac{P{Q}^2+R{S}^2}{2}}}$

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 $\mathrm{\Delta }PRQ$,
$\to \tan \theta ={\displaystyle \frac{PQ}{PR}}$ (since we know that $\tan \theta ={\displaystyle \frac{perpendicular}{base}}$)
$$\begin{gathered} \to PR={\displaystyle \frac{PQ}{\tan \theta }} \\ \Rightarrow PR=PQ\cot \theta \end{gathered}$$
Let the above equation be equation 1-
$\to PR=PQ\cot \theta$ (equation 1)
In $\mathrm{\Delta }SPR$,
$$\begin{gathered} \to \tan \left(90-\theta \right)={\displaystyle \frac{RS}{PR}} \\ \to PR={\displaystyle \frac{RS}{\cot \theta }} \\ \Rightarrow PR=RS\tan \theta \end{gathered}$$ ($\tan \left(90-\theta \right)=\cot \theta$)
Let the above equation be equation 2-
$\to PR=RS\tan \theta$ (equation 2)
From equation 1 and 2 we get-
$$\begin{gathered} \to PR=PR \\ \to PQ\cot \theta =RS\tan \theta \\ \to {\displaystyle \frac{PQ}{RS}}={\displaystyle \frac{\tan \theta }{\cot \theta }} \end{gathered}$$
$\to {\displaystyle \frac{PQ}{RS}}={\tan }^2\theta$ (${\displaystyle \frac{1}{\cot \theta }}=\tan \theta$)
$\Rightarrow \tan \theta =\sqrt{{\displaystyle \frac{PQ}{RS}}}$
Putting this value of $\tan \theta$ in equation 2, we get-
$$\begin{gathered} \to PR=RS\tan \theta \\ \to PR=RS\sqrt{{\displaystyle \frac{PQ}{RS}}} \end{gathered}$$
$\to PR=\sqrt{RS}.\sqrt{RS}{\displaystyle \frac{\sqrt{PQ}}{\sqrt{RS}}}$ (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-
$$\begin{gathered} \to 2r=PR \\ \Rightarrow 2r=\sqrt{RS\times PQ} \end{gathered}$$
Hence, A is the correct option.

Note: Remember the basic formula of $\tan \theta ={\displaystyle \frac{perpendicular}{base}}$. 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.