The shortest distance between the parabola $y^{2} = 4 x$ and the circle $x^{2} + y^{2} + 6 x - 12 y + 20 = 0$ is

4 \sqrt{2} - 5

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3 \sqrt{2} + 5

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Solution

The correct option is A $4 \sqrt{2} - 5$
Normal at a point $(m^{2}, - 2 m)$ on the parabola $y^{2} = 4 x$ is given by $y = m x - 2 m - m^{3}$ .
If this normal to the circle alos, then it will pass through center of the circle so
$6 = - 3 m - 2 m - m^{3} \Rightarrow m = - 1$
Since shortest distance between parabola and circle will occur alon common normal, shortest distance is $4 \sqrt{2} - 5$

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y^{2} = 4 x

x^{2} + y^{2} - 4 x - 2 y + 4 = 0

The equation of circle concentric with circle $x^{2} + y^{2} - 6 x + 12 y + 15$ = 0 and double its area is

x^{2} + y^{2} + 6 x + 6 y = 0

x^{2} + y^{2} - 12 x - 12 y = 0

The common chord of the circle $x^{2} + y^{2} + 4 x + 1 = 0$ and $x^{2} + y^{2} + 6 x + 2 y + 3 = 0$ is

x^{2} + y^{2} - 4 x - 6 y + 5 = 0

x^{2} + y^{2} - 2 x - 4 y - 1 = 0

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