JEE Main 2022 — Circle Question with Solution

Tangent at $R\left(x_1,y_1\right)$ when point $R$ is on the circle $x^2+y^2+2gx+2fy+c=0$ can be written as 

$x{x}_1+y{y}_1+g\left(x+x_1\right)+f\left(y+y_1\right)+c=0$.

Tangent at $O$ will be 

$-\left(x+0\right)-2\left(y+0\right)=0$ $\Rightarrow x+2y=0$

Tangent at $P$ will be 

$x\left(1+\sqrt{5}\right)+2y-\left(x+1+\sqrt{5}\right)-2\left(y+2\right)=0$

$\Rightarrow \sqrt{5}x=5+\sqrt{5}\Rightarrow x=\sqrt{5}+1$ and $y=-\frac{\sqrt{5}+1}{2}$

So $\mathrm{Q}\left(\sqrt{5}+1,-\frac{\sqrt{5}+1}{2}\right)$

Length of tangent $L=OQ=\sqrt{6+2\sqrt{5}+\frac{3+\sqrt{5}}{2}}=\sqrt{{\left(\frac{5}{2}+\frac{\sqrt{5}}{2}\right)}^2}=\frac{5+\sqrt{5}}{2}$

We know area of triangle $OPQ$ $=\frac{R{L}^3}{R^2+L^2}$

Here, radius of circle $R=\sqrt{5}$

So area$=\frac{\sqrt{5}\times {\left(\frac{5+\sqrt{5}}{2}\right)}^3}{5+{\left(\frac{5+\sqrt{5}}{2}\right)}^2}$

$=\frac{\sqrt{5}}{2}\times \frac{\left(125+75+75\sqrt{5}+5\sqrt{5}\right)}{\left(20+25+10\sqrt{5}+5\right)}=\frac{\sqrt{5}}{2}\left(\frac{200+80\sqrt{5}}{50+10\sqrt{5}}\right)$

$=\frac{\sqrt{5}}{2}\left(\frac{20+8\sqrt{5}}{5+\sqrt{5}}\right)\left(\frac{5-\sqrt{5}}{5-\sqrt{5}}\right)=\frac{\sqrt{5}}{2}\left(\frac{100-20\sqrt{5}+40\sqrt{5}-40}{20}\right)=\sqrt{5}\left(\frac{3+\sqrt{5}}{2}\right)=\frac{5+3\sqrt{5}}{2}$