Circles Part 2

Concept:The angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the radii at the centre.Explanation:Given circle: $x^{2}+y^{2}-4x-6y-3=0$Centre $O=(2,3)$, radius $r=\sqrt{4+9+3}=4$.Angle at centre $\angle AOB=\frac{\pi}{3}=60^\circ$.Thus angle between tangents $=180^\circ-60^\circ=120^\circ$.Half-angle $=60^\circ$.For external point $P(h,k)$, let $d=OP$.In right triangle with radius and tangent: $\tan 60^\circ=\frac{r}{\sqrt{d^{2}-r^{2}}}$.Here $\tan 60^\circ=\sqrt{3}$, $r=4$, and $d^{2}-r^{2}=h^{2}+k^{2}-4h-6k-3$.So $\sqrt{3}=\frac{4}{\sqrt{h^{2}+k^{2}-4h-6k-3}}$.Squaring: $3=\frac{16}{h^{2}+k^{2}-4h-6k-3}$.$\Rightarrow 3(h^{2}+k^{2}-4h-6k-3)=16$.$\Rightarrow 3h^{2}+3k^{2}-12h-18k-9=16$.$\Rightarrow 3h^{2}+3k^{2}-12h-18k-25=0$.Replace $(h,k)$ by $(x,y)$ to get locus.Answer:$3(x^{2}+y^{2})-12x-18y-25=0$ (Option C)