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 $60°$. Then, the point at which $L$ touches the line $x = 6$ is

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

The standard form of a circle is $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the center and $r$ is the radius.

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

For x: $x^2 + 4x + 4 = (x+2)^2$ (add and subtract 4)

For y: $y^2 - 6y + 9 = (y-3)^2$ (add and subtract 9)

$(x+2)^2 + (y-3)^2 - 4 - 9 - 3 = 0$

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

Center $= (-2, 3)$ and Radius $= \sqrt{16} = 4$

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When two tangents are drawn from an external point P to a circle, the tangents have equal length and the angle between the center-to-point line and each tangent is the same.

If the angle between the two tangents is $60°$, then each tangent makes an angle of $30°$ with the line joining the external point to the center.

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In the right triangle formed by:

A $=$ point of tangency

C $=$ center of circle $(-2, 3)$

P $=$ external point

We have:

$\angle CAP = 90°$ (radius perpendicular to tangent)

$\angle ACP = 30°$ (half of the $60°$ angle between tangents)

$AC = 4$ (radius)

Using trigonometry in right triangle ACP:

$\sin(30°) = \tfrac{AC}{CP} = \tfrac{4}{CP}$

Since $\sin(30°) = \tfrac{1}{2}$:

$\tfrac{1}{2} = \tfrac{4}{CP}$

Therefore: $CP = 8$

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The locus L consists of all points P that are at distance 8 from the center $(-2, 3)$.

Locus equation: $(x+2)^2 + (y-3)^2 = 64$

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We substitute $x = 6$ into the locus equation:

$(6+2)^2 + (y-3)^2 = 64$

$64 + (y-3)^2 = 64$

$(y-3)^2 = 0$

$y = 3$

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The point where L touches the line $x = 6$ is $(6, 3)$.

The point $(6, 3)$ is exactly 8 units away from the center $(-2, 3)$, confirming our geometric analysis. From this point, tangents drawn to the circle will indeed make a $60°$ angle with each other.