For $x^{2}+y^{2}-2 \sqrt{2} x-6 \sqrt{2} y+14=0$ 

$\text { Radius }=\sqrt{(\sqrt{2})^{2}+(3 \sqrt{2})^{2}-14}=\sqrt{6}$

$\Rightarrow$ Diameter $=2 \sqrt{6}$ 

If this diameter is chord to 

$\begin{aligned} &(x-2 \sqrt{2})^{2}+(y-2 \sqrt{2})^{2}=r^{2} \text { then } \\ &\Rightarrow r^{2}=6+\left(\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}}\right)^{2} \\ &\Rightarrow r^{2}=6+4=10 \\ &\Rightarrow r^{2}=10 \end{aligned}$