Maximum Distance from Circle Point to Parabola Vertex

To find the maximum distance of a point $P$ on the circle from the vertex of the parabola, we first determine the center and radius of the circle, as well as the coordinates of the vertex of the parabola.

Step 1: Analyze the given circle

The equation of the circle is: $x^2 + y^2 - 6x - 8y + 21 = 0$

Completing the square for both $x$ and $y$: $(x^2 - 6x + 9) + (y^2 - 8y + 16) - 9 - 16 + 21 = 0$ $(x - 3)^2 + (y - 4)^2 - 4 = 0$ $(x - 3)^2 + (y - 4)^2 = 2^2$

Thus, the center of the circle is $C(3, 4)$ and its radius is $r = 2$.

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Step 2: Determine the vertex of the parabola

The equation of the parabola is: $x^2 + 6x + y + 13 = 0$

Completing the square for $x$: $(x^2 + 6x + 9) + y + 4 = 0$ $(x + 3)^2 = -(y + 4)$

This is a standard parabola opening downwards with its vertex at $V(-3, -4)$.

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Step 3: Calculate the maximum distance

The distance between any point $P$ on a circle centered at $C$ with radius $r$ and a fixed external point $V$ satisfies: $d_{\text{max}} = CV + r$

First, calculate the distance $CV$ between the center of the circle $C(3, 4)$ and the vertex of the parabola $V(-3, -4)$: $CV = \sqrt{(3 - (-3))^2 + (4 - (-4))^2}$ $CV = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$

Now, adding the radius $r = 2$ of the circle: $d_{\text{max}} = 10 + 2 = 12$

Thus, the maximum distance of $P$ from the vertex of the parabola is 12.

Correct Option: C