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

$$x^2+y^2-4x-6y-3$$= 0

$$(x-2)^2+(y-3)^2=16$$

So the center is (O(2,3)) and radius (r = 4).

Let (A) and (B) be points on the circle such that
($$\angle AOB=\pi/3=60^{\circ}$$).

Tangents at (A) and (B) meet at a point (P).

A standard result in circle geometry:

The intersection point of tangents at two points subtending angle (\theta) at the center lies on a circle centered at (O) with radius
OP =$$\frac{r}{\cos(\theta/2)}$$

OP =$$\frac{4}{\cos(30^{\circ})}=\frac{4}{\sqrt{3}/2}=\frac{8}{\sqrt{3}}$$

So (P) moves on a circle centered at (O(2,3)) with radius $$(8/\sqrt{3})$$.

$$(x-2)^2+(y-3)^2=\frac{64}{3}$$