Let $$P$$ moving point on the circle $$x^2 + y^2 - 6x - 8y + 21 = 0$$. Then,the maximum distance of $$P$$ from the vertex of the parabola $$x^2 + 6x + y + 13 = 0$$ is :

To find the maximum distance of point $$P$$ (on the circle) from the vertex of the parabola, we need to find the coordinates of the circle's center, its radius, and the vertex of the parabola.

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

We complete the square to find the center $$(h, k)$$ and radius $$r$$:

$$(x^2 - 6x + 9) + (y^2 - 8y + 16) = -21 + 9 + 16$$

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

• Center ($$C$$): $$(3, 4)$$
• Radius ($$r$$): $$\sqrt{4} = 2$$
• Vertex ($$V$$): $$(-3, -4)$$

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

Rearrange to find the vertex:

$$y = -x^2 - 6x - 13$$

$$y = -(x^2 + 6x + 9) - 13 + 9$$

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

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

For any point $$P$$ on a circle, the distance to an external point $$V$$ is maximized when $$P$$ lies on the line passing through the center $$C$$ and the point $$V$$, specifically on the "far side" of the circle.

The maximum distance $$d_{max}$$ is given by:

$$d_{max} = CV + r$$

First, calculate the distance between the center $$C(3, 4)$$ and the vertex $$V(-3, -4)$$ using the distance formula:

$$CV = \sqrt{(3 - (-3))^2 + (4 - (-4))^2}$$

$$CV = \sqrt{6^2 + 8^2}$$

$$CV = \sqrt{36 + 64}$$

$$CV = \sqrt{100} = 10$$

Now, add the radius of the circle:

$$d_{max} = 10 + 2 = 12$$

Conclusion:

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

Correct Option: C