Question

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

• A. $\sqrt{5}$
• B. $\sqrt{21}$
• C. $\sqrt{28}-\sqrt{5}$
• D. $\sqrt{21}-\sqrt{5}$

Answer 1

Answer: (A)

Centre and radius of the given circle is $P(6,0)$ and $\sqrt{5}$, respectively.
Now minimum distance between two curves always occurs along a line which normal to both the curves.
Equation of normal to $y^2=4x$ at $\left(t^2,2t\right)$ is
$y=-tx+2t+t^3$

If it is normal to circle also, then it must pass though $(6,0)$.
$\therefore 0=t^3-4t$
$\Rightarrow t=0$ or $t=\pm 2$
$\Rightarrow A(4,4)$ and $(4,-4)$
$\Rightarrow PA=PC=\sqrt{20}=2\sqrt{5}$
$\Rightarrow$ Required minimum distance $2\sqrt{5}-\sqrt{5}=\sqrt{5}$