Thank you for reporting, we will resolve it shortly
Q.
If the tangents drawn at the point $O (0,0)$ and $P (1+\sqrt{5}, 2)$ on the circle $x ^{2}+ y ^{2}-2 x -4 y =0$ intersect at the point $Q$, then the area of the triangle OPQ is equal to
Tangent at $O$
$-( x +0)-2( y +0)=0$
$\Rightarrow x +2 y =0$
Tangent at $P$
$x (1+\sqrt{5})+ y \cdot 2-( x +1+\sqrt{5})-2( y +2=0)$
Put $x=-2 y$
$-2 y (1+\sqrt{5})+2 y +2 y -1-\sqrt{5}-2 y -4=0$
$-2 \sqrt{5} y =5+\sqrt{5}$
$\Rightarrow y =\left(\frac{\sqrt{5}+1}{2}\right)$
$Q \left(\sqrt{5}+1,-\frac{\sqrt{5}+1}{2}\right)$
Length of tangent $OQ =\frac{5+\sqrt{5}}{2}$
Area $=\frac{ RL ^{3}}{ R ^{2}+ L ^{2}}$
$R =\sqrt{5}$
$=\frac{\sqrt{5} \times\left(\frac{5+\sqrt{5}}{2}\right)^{3}}{5+\left(\frac{5+\sqrt{5}}{2}\right)^{2}}$
$=\frac{\sqrt{5}}{2} \times \frac{4 \times(125+75+75 \sqrt{5}+5 \sqrt{5})}{(20+25+10 \sqrt{5}+5)}$
$=\frac{5+3 \sqrt{5}}{2}$