Let P be the point on the parabola, y^2 = 8x which is at a minimum distance

• Coordinate Geometry

Let P be the point on the parabola, y² = 8x which is at a minimum distance from the centre C of the circle, x² + (y + 6)² = 1. Then the equation of the circle, passing through C and having its centre at P is:

1. x² + y² – x + 4y – 12 = 0
2. x² + y² – x/4 + 2y – 24 = 0
3. x² + y² – 4x + 9y + 18 = 0
4. x² + y² – 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: y² = 4ax = 8x

a = 2

Let P(at², 2at) = (2t², 4t)

Equation of normal to parabola is

y = –tx + 2at + at³

y = –tx + 4t + 2t³

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

–6 = 4t + 2t³

t³ + 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.

r² = (2-0)² + (-4+6)² = 4+4 = 8

Equation of required circle:

(x – 2)² + (y + 4)² = 8

x² + y² – 4x + 8y + 12 = 0

The correct option is D.