Question

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

Answer 1

Intersection points

When $x>0$ , $x^2+x+4=0$ , $y=x$

$x=\frac{-1+\sqrt{17}}{2}=y$

When $x<0$ , $y=-x$ , $x^2-x-4=0$

$\frac{1-\sqrt{17}}{2}$

$y=\frac{\sqrt{17}-1}{2}$

Due to symmetry

$2\left[{\int }_0^{\sqrt{17}-1/2}(4-x^2)dx-\frac{1}{2}{\left(\frac{\sqrt{17}-1}{2}\right)}^2\right]$

$2[\frac{7\sqrt{17}+1}{6}-\frac{1}{2}\times \frac{1}{4}[14+1-2\sqrt{17}\left]\right]$

$2\left[\frac{28\sqrt{17}+4-54+6\sqrt{17}}{24}\right]$

$\Rightarrow$ $2\left[\frac{34\sqrt{17}-50}{24}\right]=\frac{34\sqrt{17}-50}{12}$ units