Minimum distance between curves y^2 =4x and x^2 +y^2 −12x+31=0 is ?


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Question Number 57578 by rahul 19 last updated on 07/Apr/19

$M i n i m u m d i s t a n c e b e t w e e n c u r v e s$ $y^{2} = 4 x a n d x^{2} + y^{2} - 12 x + 31 = 0 i s ?$
	Answered by mr W last updated on 07/Apr/19

	$x^{2} + y^{2} - 12 x + 31 = 0$ ${(x - 6)}^{2} + y^{2} = {(\sqrt{5})}^{2}$ $\Rightarrow c i r c l e w i t h r a d i u s \sqrt{5} a n d c e n t e r (6, 0) .$ $e q n . o f c i r c l e w i t h r a d i u s r a n d c e n t e r (6, 0) :$ ${(x - 6)}^{2} + y^{2} = r^{2}$ $s u c h t h a t t h e c i r c l e t o u c h e s t h e c u r v e$ $y^{2} = 4 x$ $\Rightarrow {(x - 6)}^{2} + 4 x = r^{2}$ $\Rightarrow x^{2} - 8 x + (36 - r^{2}) = 0$ $\Rightarrow Δ = 64 - 4 (36 - r^{2}) = 0$ $\Rightarrow r = 2 \sqrt{5}$ $M i n i m u m d i s t a n c e b e t w e e n c u r v e s$ $y^{2} = 4 x a n d x^{2} + y^{2} - 12 x + 31 = 0 i s$ $2 \sqrt{5} - \sqrt{5} = \sqrt{5}$
	Commented by mr W last updated on 07/Apr/19

	Commented by rahul 19 last updated on 08/Apr/19
	thank you sir.
	Answered by MJS last updated on 07/Apr/19

	$the center of the circle is (\begin{matrix} 6 \\ 0 \end{matrix}); r = \sqrt{5}$ $\Rightarrow the point of the parabola with minimum$ $distance from this center is the same as the$ $point with the minimum distance of the$ $circle .$ $∣ [\begin{matrix} x \\ 2 \sqrt{x} \end{matrix}] - [\begin{matrix} 6 \\ 0 \end{matrix}] ∣^{2} = x^{2} - 8 x + 36$ $\frac{d}{d x} [x^{2} - 8 x + 36] = 2 x - 8 = 0 \Rightarrow x = 4$ $distance = \sqrt{x^{2} - 8 x + 36} - \sqrt{5} = \sqrt{20} - \sqrt{5} = \sqrt{5}$
	Commented by rahul 19 last updated on 08/Apr/19
	thank you sir.
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