What is the shortest distance between the circle x^2 + (y+6)^2 = 1 and the parabola y^2 = 8x?

Question

What is the shortest distance between circle A defined by (x − 5)2 + (y − 4)2 = 4 and circle B defined by (x − 1)2 + (y − 1)2 = 1? There is some point P on circle A closest to B, and there is another point Q on circle B closest to A. What are the coordinates of P and Q?

Answer 1

Minimum distance between any two curves lie along common normal
So if we find the coordinate where common normal touches the parabola, then that coordinate will be closest to circle
y^2 = 8x
x^2 + (y+6)^2 = 1
Equation normal to parabola:
y = mx - 2(2)m-2m^3
where m is slope of normal
y = mx - 4m-2m^3
So this common normal will pass through the center of circle also
Center (0, -6)
So, -6 = m(0) - 4m-2m^3
m^3 + 2m - 3 = 0
m = 1 is one solution
y = x - 6
Put this in parabola
y^2 = 8(y+6)
y^2 - 8y - 48 = 0
y = -4, 12
x = 2, 18
(2, -4) is the closest point