Let C be the circle $$x^{2} + y^{2} + 4x - 6y - 3 = 0$$ and L be the locus of the point of intersection of a pair of tangents to C with the angle between the two tangents equal to $$60^{\circ}$$. Then, the point at which L touches the line $$x$$ = 6 is

Given equation of circle = $$x^{2} + y^{2} + 4x - 6y - 3 = 0$$

Center of the circle is (-2,3) and radius of the circle = $$\sqrt{\ g^2+f^2-c}=\sqrt{\ 4+9+3}=4$$

Let us assume the point of the intersection of the tangents is ( h,k)

The angle made by the line joining (h,k) to the centre makes an angle of 30 degrees with the tangent, and sin(30) will be the ratio of the radius and the distance between the center and (h,k)

=> $$\sin\left(30\right)=\dfrac{4}{\sqrt{\left(\ h+2\right)^2+\left(k-3\right)^2}}$$

Squaring on both sides:

$$\dfrac{1}{4}=\dfrac{16}{\left(h+2\right)^2+\left(k-3\right)^2}$$

=> $$\left(h+2\right)^2+\left(k-3\right)^2=64$$

When x = 6 => h = 6 => $$64+\left(k-3\right)^2=64$$ => k = 3.

=> required point is (6,3)