The circle $x=5 \cos \theta, y=5 \sin \theta$ is bounded by the rectangle formed by the lines $x \pm 6=0$ and $y \pm 6=0$. The area of the triangle that lies inside the rectangle which is formed by the tangent at $P\left(\frac{2 \pi}{3}\right)$ to the circle with two of the above given lines is

• Correct answer
  A

  $\frac{62-24 \sqrt{3}}{\sqrt{3}}$
• B

  $\frac{1}{2}(6 \sqrt{3}-4)^2$
• C

  $48+\sqrt{3}$
• D

  $\frac{1}{2}\left(\frac{6 \sqrt{3}-4}{\sqrt{3}}\right)^2$

Solution

Coordinate of $P$ will be $\left(5 \cos \frac{2 \pi}{3}, 5 \sin \frac{2 \pi}{3}\right)$
$$ \begin{aligned} & =\left(-\frac{5}{2}, \frac{5 \sqrt{3}}{2}\right) \\ \text { Now, }\left(\frac{d y}{d x}\right)_{\theta=\frac{2 \pi}{3}} & =-\cot \left(\frac{2 \pi}{3}\right)=\frac{1}{\sqrt{3}} \end{aligned} $$ $\therefore$ Equation of line $P Q \Rightarrow y-\frac{5 \sqrt{3}}{2}=\frac{1}{\sqrt{3}}\left(x+\frac{5}{2}\right)$
To find coordinate of $Q$, we put $y=6$ in Eq. (i), $$ \begin{aligned} & x-6 \sqrt{3}+10=0 \Rightarrow x=6 \sqrt{3}-10 \\ & \therefore \quad Q=(6 \sqrt{3}-10,6) \\ & \text { Similarly, } R=\left(-6, \frac{4}{\sqrt{3}}\right) \end{aligned} $$ From figure, $S=(-6,6)$ $\therefore$ Required area $=$ Area of $\triangle R S Q$ $=\frac{1}{2}\left|\begin{array}{ccc}-6 & \frac{4}{\sqrt{3}} & 1 \\ 6 \sqrt{3}-10 & 6 & 1 \\ -6 & 6 & 1\end{array}\right|$ $$ \begin{aligned} & =\frac{1}{2}\left|\left[0-\frac{4}{\sqrt{3}}(6 \sqrt{3}-10+6)+1(36 \sqrt{3}-60+36)\right]\right| \\ & =\frac{1}{2}\left|-24+\frac{16}{\sqrt{3}}+36 \sqrt{3}-24\right|=\left|\frac{8}{\sqrt{3}}+18 \sqrt{3}-24\right| \\ & =\left|\frac{8+54-24 \sqrt{3}}{\sqrt{3}}\right|=\frac{62-24 \sqrt{3}}{\sqrt{3}} \end{aligned} $$

Hence, the correct answer is option: 'A'.