Find the minimum distance between } the curves }y^2=4x and }x^2+y^2-12x+31=0 . } solve it by parametric form of equation of normal of a parabola

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y^{2} = 4 x

x^{2} + y^{2} - 12 x + 31 = 0

Minimum distance between the curve $y^{2} = 4 x$ and $x^{2} + y^{2} - 12 x + 31 = 0$ is

x^{2} + y^{2} + 4 x + 16 y + 66 = 0

y^{2} = 8 x

y^{2} = 4 x

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

S_{1} \equiv x^{2} + y^{2} - 4 y - 5 = 0

S_{2} \equiv x^{2} + y^{2} + 12 x + 4 y + 31 = 0

S_{2} \equiv x^{2} + y^{2} + 6 x + 12 y + 36 = 0

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