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. $\frac{\sqrt{17}-\sqrt{2}}{2}$

Answer 1

The correct option is D $\frac{\sqrt{17}-\sqrt{2}}{2}$


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.