Question

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

• A. $\sqrt{2}$
• B. $\sqrt{5}$
• C. $2\sqrt{2}$
• D. $2\sqrt{5}$

Answer 1

The correct option is B

$\sqrt{5}$

Shortest distance between the two curves will be along common normal.

$\therefore$ Equation of normal at $(t^2,2t)$ is $y=-tx+2t+t^3$

But it passes through centre (6, 0) of circle

$\therefore t^3-4t=0\Rightarrow t=0$ or $t=\pm 2$

$\therefore A=(4,4),C=(4,-4)$

$\Rightarrow AD=AP–PD$

$=2\sqrt{5}-\sqrt{5}=\sqrt{5}$