JEE Main · 2017 · 02 Apr 2017 · Offline · Application of Derivatives

By symmetry the centre lies on the $y$-axis at $(0,\lambda )$. Since the circle touches the lines $y=|x|$, its radius equals the distance from the centre to $x-y=0$, giving $r=\frac{\lambda }{\sqrt{2}}$. The smallest such circle that also touches the parabola $y=4-x^2$ makes contact at the vertex $(0,4)$, where the parabola has a horizontal tangent matching the top of the circle, so $\lambda +r=4$. Substituting $\lambda =r\sqrt{2}$ gives $r\sqrt{2}+r=4$, i.e. $r(\sqrt{2}+1)=4$, so $r=\frac{4}{\sqrt{2}+1}=4(\sqrt{2}-1)$.