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

Question

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

Accepted Answer

By symmetry, take the circle centred at $(0,k)$ on the $y$-axis, touching $y=x$: $\dfrac{|{-k}|}{\sqrt2}=r\Rightarrow k=r\sqrt2$.

Touching $y=4-x^2$: substituting $x^2=4-y$ into $x^2+(y-k)^2=k^2/2$ and setting the discriminant to zero:
$$2k^2+4k-15=0\ \Longrightarrow\ k=\frac{-2+\sqrt{34}}2\ \Longrightarrow\ r=\frac k{\sqrt2}=\frac{-2+\sqrt{34}}{2\sqrt2},$$

which does not match any of options A--D.

Solutions · 1