Let PQ and MN be two straight lines touching the circle x^2+y^2-4 x-6 y-3=0 at the points …

Question

Let PQ and MN be two straight lines touching the circle $x^2+y^2-4 x-6 y-3=0$ at the points $A$ and $B$ respectively. Let $O$ be the centre of the circle and $\angle A O B=\pi / 3$. Then the locus of the point of intersection of the lines P...

Accepted Answer

$\begin{gathered} {x}^{2}+{y}^{2}-14x-6y-3=0 \ 	an {60}^{\circ }=\frac{4}{\sqrt{{h}^{2}+{k}^{2}-4h-6k-3}} \end{gathered}$

squaring both side

$3({h}^{2}+{k}^{2}-4h-6k-3)=16$

To get locus replace $h,k$ by $x$ and $y$;

$\begin{gathered} 3{x}^{2}+3{y}^{2}-12x-18y-9-16=0 \ 3{x}^{2}+3{y}^{2}-12x-18y-25=0 \end{gathered}$

Solutions · 1