Let P be a 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 equal to:

Step-by-step solution

The equation of the given circle is x^2 + y^2 - 6x - 8y + 21 = 0 . The center of the circle is C(3, 4) and its radius is r = 3^2 + 4^2 - 21 = 4 = 2 . The equation of the given parabola is x^2 + 6x + y + 13 = 0 . Rewriting the equation by completing the square, we get (x + 3)^2 = -(y + 4) . The vertex of the parabola is V(-3, -4) . The maximum distance of a moving point P on the circle from the vertex V is given by CV + r . The distance between the center C(3, 4) and the vertex V(-3, -4) is CV = (3 - (-3))^2 + (4 -