
$\begin{aligned}
& C: x^2+y^2=10 \
& A N=\frac{M N}{2}=1 \
& \therefore \text { In } \triangle O A N \rightarrow(O N)^2=(O A)^2+(A N)^2 \
& 10=(O A)^2+1 \rightarrow O A=3
\end{aligned}$
Perpendicular distance of center from PQ=2∣0+0−2∣=2
Perpendicular distance between MN and $\begin{aligned}
& \mathrm{PQ}=\mathrm{OA}+\sqrt{2} \text { or }|\mathrm{OA}-\sqrt{2}| \
& =3+\sqrt{2} \text { or } 3-\sqrt{2}
\end{aligned}$