Question

The radius of a circle, having minimum area, which touches the curve $y=4-x^2$ and the lines, $y=|x|$ is.

• A. $2(\sqrt{2}+1)$
• B. $2(\sqrt{2}-1)$
• C. $4(\sqrt{2}-1)$
• D. $4(\sqrt{2}+1)$

Answer 1

The correct option is C $4(\sqrt{2}-1)$

Let the centre be $(0,k)$

Now, we know the radius as $r=4-k$

Now the radius will be equal to the perpendicular distance from the lines.

$4-k=\left|\frac{k}{\sqrt{2}}\right|$

$\Rightarrow k=\frac{4\sqrt{2}}{1+\sqrt{2}}$

Radius, $r=4-k=4-\frac{4\sqrt{2}}{\sqrt{2}+1}$

$r=\frac{4}{\sqrt{2}+1}$

$\Rightarrow r=4(\sqrt{2}-1)$