The circle will have centre on $x=y$ line since parabolas are symmetric about $y=x$ line

The slope of tangent at closest point $y^2=x-2$

$\Rightarrow 2 y\left(\frac{d y}{d x}\right)=1 \Rightarrow y=\frac{1}{2} \Rightarrow$ point will be $\left(\frac{9}{4}, \frac{1}{2}\right)$

Similarly, on $x^2=y-2 \Rightarrow 2 x=\frac{d y}{d x}=1 \Rightarrow\left(\frac{1}{2}, \frac{9}{4}\right)$

$$\begin{aligned} & 2 r=\sqrt{\left(\frac{9}{4}-\frac{1}{2}\right)^2+\left(\frac{1}{2}-\frac{9}{4}\right)^2}=\sqrt{2} \cdot\left|\frac{9}{4}-\frac{1}{2}\right| \\ & =\frac{14}{8} \sqrt{2}=\frac{7 \sqrt{2}}{4} \\ & \Rightarrow r=\frac{7 \sqrt{2}}{8} \end{aligned}$$