CAT 2023 Slot 1 Solved Paper

Given equation of circle $={\mathit{x}}^2+{\mathit{y}}^2+4\mathit{x}-6\mathit{y}-3=0$
Center of the circle is $\left(-2,3\right)$ and radius of the circle $=\sqrt{{\mathit{g}}^2+{\mathit{f}}^2-\mathit{c}}=\sqrt{4+9+3}=4$
Let us assume the point of the intersection of the tangents is $\left(\mathit{h},\mathit{k}\right)$
The angle made by the line joining $\left(\mathit{h},\mathit{k}\right)$ to the centre makes an angle of 30 degrees with the tangent, and $\mathit{s}\mathit{i}\mathit{n}\left(30\right)$ will be the ratio of the radius and the distance between the center and $\left(\mathit{h},\mathit{k}\right)$
$\Rightarrow \mathit{s}\mathit{i}\mathit{n}\left(30\right)=\frac{4}{\sqrt{{\left(\mathit{h}+2\right)}^2+{\left(\mathit{k}-3\right)}^2}}$
Squaring on both sides:
$\frac{1}{4}=\frac{16}{{\left(\mathit{h}+2\right)}^2+{\left(\mathit{k}-3\right)}^2}$
$\Rightarrow {\left(\mathit{h}+2\right)}^2+{\left(\mathit{k}-3\right)}^2=64$
When $\mathrm{x}=6\Rightarrow \mathrm{h}=6\Rightarrow 64+{\left(\mathit{k}-3\right)}^2=64\Rightarrow \mathrm{k}=3.$
⇒ required point is $\left(6,3\right)$