Question

The minimum area of circle which touches the parabolas $y=\frac{1}{2}\left(x^2+5\right)$ and $x=\frac{1}{2}\left(y^2+5\right)$ is

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

Answer 1

Answer: (B)

Parabolas, $y=\frac{1}{2}\left(x^2+5\right)$ and $x=\frac{1}{2}\left(y^2+5\right)$ are symmetrical about the line $y=x$
$\therefore$ tangent at $A$ is parallel to $y=x$
$y=\frac{1}{2}\left(x^2+5\right)$
$\frac{dy}{dx}=\frac{1}{2}(2x)=1\Rightarrow x=1$
co-ordinate of $A$ is $(1,3)$
$\therefore$ co-ordinate of $B$ is $(3,1)$
Radius of circle $(r)=\frac{1}{2}(AB)=\frac{1}{2}\sqrt{4+4}=\sqrt{2}$
Area of circle $=\pi r^2=2\pi$