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 PQ and MN is :

Question

Accepted Answer

C. 1. **Set up the tangent geometry.** Let $T(h,k)$ be the point of intersection of the tangents $PQ$ and $MN$. Since $O$ is the centre and $OA\perp TA$, $OB\perp TB$, the quadrilateral $OATB$ has two right angles, so
$$
\angle AOB+\angle ATB=\pi
$$
Given $\angle AOB=\frac{\pi}{3}$, this gives $\angle ATB=\frac{2\pi}{3}$, so each half-angle at $T$ is
$$
\angle ATO=\frac{\pi}{3}
$$

2. **Relate this angle to the tangent length.** In right triangle $OAT$, with $OA=r$ (the radius) and $AT=L$ (the tangent length from $T$),
$$
	an(\angle ATO)=\frac{OA}{AT}=\frac{r}{L}
$$

3. **Find the radius of the given circle.** For
$$
x^2+y^2-4x-6y-3=0
$$
the radius is
$$
r=\sqrt{4+9+3}=4
$$

4. **Express the tangent length from $T(h,k)$.** The length of the tangent from an external point $(h,k)$ to a circle $x^2+y^2+2gx+2fy+c=0$ is $\sqrt{h^2+k^2+2gh+2fk+c}$, so here
$$
L=\sqrt{h^2+k^2-4h-6k-3}
$$

5. **Substitute into the relation from step 2.**
$$
	an\frac{\pi}{3}=\frac{4}{\sqrt{h^2+k^2-4h-6k-3}}
$$

6. **Square both sides and simplify.** Since $	an\frac{\pi}{3}=\sqrt3$,
$$
3\left(h^2+k^2-4h-6k-3ight)=16
$$

7. **Rearrange into the locus equation.**
$$
3h^2+3k^2-12h-18k-9-16=0 \implies 3h^2+3k^2-12h-18k-25=0
$$
Replacing $(h,k)$ with $(x,y)$,
$$
3(x^2+y^2)-12x-18y-25=0
$$

Hence, the locus of the point of intersection of $PQ$ and $MN$ is $3(x^2+y^2)-12x-18y-25=0$, so the answer is **Option C**.