Let P be the point on the parabola, $y^{2} = 8 x$ 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:

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Solution

The correct option is D

Let the normal of parabola be

$y = m x - 4 m - 2 m^{3}$

(0, – 6) lies on it

$∴ - 6 = - 4 m - 2 m^{3}$

$\Rightarrow m^{3} + 2 m - 3 = 0$

(m – 1) (m2 + m + 3) = 0

m = 1

$∴ P o i n t P : (2 m^{2}, - 4 m)$

= (2, - 4)

$∴ Equation of circle is$

$(x - 2)^{2} + (y + 4)^{2} = (4 + 4) \Rightarrow x^{2} + y^{2} - 4 x + 8 y + 12 = 0$

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y^{2} = 8 x

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