Let P be the point on the parabola, y^{2} = 8x which is at a minimum distance from the centre C of the circle, x^{2} + (y + 6)^{2} = 1. Then the equation of the circle, passing through C and having its centre at P is:

- x
^{2}+ y^{2}– x + 4y – 12 = 0 - x
^{2}+ y^{2}– x/4 + 2y – 24 = 0 - x
^{2}+ y^{2}– 4x + 9y + 18 = 0 - x
^{2}+ y^{2}– 4x + 8y + 12 = 0

**Solution**

For minimum distance from the centre of circle to the parabola at point P, the line must be normal to the parabola at P.

Equation of the parabola: y^{2} = 4ax = 8x

a = 2

Let P(at^{2}, 2at) = (2t^{2}, 4t)

Equation of normal to parabola is

y = –tx + 2at + at^{3}

y = –tx + 4t + 2t^{3}

It passes through centre of circle C(0, –6)

–6 = 4t + 2t^{3}

t^{3} + 2t + 3 = 0

t = –1

Hence P is (2, –4), which is centre of required circle.

Radius of required circle = Distance between C and P.

r^{2} = (2-0)^{2} + (-4+6)^{2} = 4+4 = 8

Equation of required circle:

(x – 2)^{2} + (y + 4)^{2} = 8

x^{2} + y^{2} – 4x + 8y + 12 = 0

**The correct option is D.**