Locus of Tangents Intersection to Circle

Circle $C$: Center $O(-2, 3)$, Radius $R = \sqrt{(-2)^2 + 3^2 - (-3)} = \sqrt{16} = 4$.

Let $P$ be a point on the locus $L$. The angle between tangents from $P$ to circle $C$ is $60^\circ$, so the angle between $OP$ and a tangent line is $30^\circ$. $\sin(30^\circ) = \frac{R}{OP} \implies \frac{1}{2} = \frac{4}{OP} \implies OP = 8$

Thus, $L$ is a concentric circle with center $(-2, 3)$ and radius $8$. Equation of $L$: $(x + 2)^2 + (y - 3)^2 = 8^2 = 64$.

For $x = 6$: $(6 + 2)^2 + (y - 3)^2 = 64 \implies 64 + (y - 3)^2 = 64 \implies y - 3 = 0 \implies y = 3$

So $L$ touches $x = 6$ at the point $(6, 3)$.