The radius of the smallest circle which touches the parabolas $y=x^2+2$ and $x=y^2+2$ is

Question

Aicharya · Accepted Answer

The given parabolas are symmetric about the line $y=x$

Tangents at $\mathrm{A} \& \mathrm{~B}$ must be parallel to $\mathrm{y}=\mathrm{x}$ line, so slope of the tangents $=1$
$\left(\frac{d y}{d x}\right)_{\min A}=1=\left(\frac{d y}{d x}\right)_{\min B}$
For point $\mathrm{B}, \mathrm{y}=\mathrm{x}^2+2$
$\begin{gathered}
\frac{d y}{d x}=2 x=1 \\ x=\frac{1}{2} \Rightarrow y=\frac{9}{4} \\ \therefore \text { Point } B=\left(\frac{1}{2}, \frac{9}{4}\right) \Rightarrow \text { Point } \mathrm{A}=\left(\frac{9}{4}, \frac{1}{2}\right) \\ \mathrm{AB}=\sqrt{\left(\frac{1}{2}-\frac{9}{4}\right)^2+\left(\frac{9}{4}-\frac{1}{2}\right)^2} \\ =\sqrt{\frac{98}{16}}=\frac{7 \sqrt{2}}{4}
\end{gathered}$
Radius $==\frac{7 \sqrt{2}}{8}$ ,

The radius of the smallest circle which touches the parabolas $y=x^2+2$ and $x=y^2+2$ is

Solution