JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Parabola

The parabolas $y=x^2+2$ and $x=y^2+2$ are mirror images in the line $y=x$, so the smallest circle touching both is centred on $y=x$ and contacts each where the tangent has slope $1$. For $y=x^2+2$, $\frac{dy}{dx}=2x=1$ gives $x=\frac{1}{2},y=\frac{9}{4}$, the point $\left(\frac{1}{2},\frac{9}{4}\right)$; by symmetry the other contact is $\left(\frac{9}{4},\frac{1}{2}\right)$. The distance between them is $\sqrt{2}\left(\frac{9}{4}-\frac{1}{2}\right)=\sqrt{2}\cdot \frac{7}{4}=\frac{7\sqrt{2}}{4}$, which is the diameter. The radius of the smallest circle is $\frac{7\sqrt{2}}{8}$.