Single Choice

Minimum distance between the curves $$y^2=4x\;\&\;x^2+y^2-12x+31=0$$ is

A$$\sqrt{21}$$
B$$\sqrt{26}-\sqrt{5}$$
C$$\sqrt{20}-\sqrt{5}$$
Correct Answer
D$$\sqrt{21}-\sqrt{5}$$

Solution

Let normal to the parabola $$y^2=4x$$ at point $$P(t^2,2t)$$ be $$y+tx=2t+t^3$$
for shortest distance it must pass through centre of the given circle i.e $$C(6,\,0)$$
$$\Rightarrow 6t=2t+t^3$$.
$$\Rightarrow t=2$$
Thus point P is $$(4,4)$$
Hence shortest distance $$=CP -r = \sqrt{4+16}-\sqrt{5}=\sqrt{20}-\sqrt{5}$$

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