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 $r$ be the radius of the circle.
$\text{Then}r=OM=\frac{k}{\sqrt{2}}$
$\Rightarrow k=\sqrt{2}r$
$\text{Also,}r=ON=4-k$
$\Rightarrow k=4-r$
$\therefore \sqrt{2}r=4-r$
$\Rightarrow r=\frac{4}{\sqrt{2}+1}=4(\sqrt{2}-1)$

Note: The creator of the question has given the options according to the first figure. But the actual figure should be the figure drawn below as the two graphs only touch each other :


Due to symmetry of the graph, centre of the circle must be on y-axis.
Let the centre be $O(0,k).$
Then the radius of the circle is equal to the length of the perpendicular from $O$ to $y=x.$
$\therefore r=OR=\frac{|k|}{\sqrt{2}}$
Then, equation of the circle is $x^2+(y-k{)}^2=\frac{k^2}{2}\cdots (1)$
$\text{Also,}y=4-x^2\cdots \left(2\right)$
Solving $\left(1\right)\&\left(2\right),$
$4-y+(y-k{)}^2=\frac{k^2}{2}$
$\Rightarrow y^2-(2k+1)y+(\frac{k^2}{2}+4)=0\cdots \left(3\right)$
y-coordinate of $P$ and $Q$ is same.
$\Rightarrow D=0$ for equation $\left(3\right).$
$\Rightarrow (2k+1{)}^2-4(\frac{k^2}{2}+4)=0$
$\Rightarrow k=\frac{-4\pm \sqrt{136}}{4}$
$\therefore r=\frac{|-4\pm \sqrt{136}|}{4\sqrt{2}}$
$\therefore r_{min}=\frac{\sqrt{17}-\sqrt{2}}{2}$
But, from the given options $r=4(\sqrt{2}-1)$ is minimum.