If the tangents drawn at the point O(0, 0) and P(1 + 5 , 2) on the...

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

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

Tangent at $P$ will be

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

$⇒ \sqrt{5}x = 5 + \sqrt{5} ⇒ x = \sqrt{5} + 1$ and $y = - \dfrac{\sqrt{5} + 1}{2}$

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

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

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

Here, radius of circle $R = \sqrt{5}$

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

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

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