If one of the diameters of the circle x^{2}+y | vedclass

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(D) Given the first circle $S: x^{2}+y^{2}-2 \sqrt{2} x-6 \sqrt{2} y+14=0$.The center $C$ of this circle is $(\sqrt{2}, 3 \sqrt{2})$ and its radius $r

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$10$

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(D) Given the first circle $S: x^{2}+y^{2}-2 \sqrt{2} x-6 \sqrt{2} y+14=0$.The center $C$ of this circle is $(\sqrt{2}, 3 \sqrt{2})$ and its radius $r_{1} = \sqrt{(\sqrt{2})^{2} + (3\sqrt{2})^{2} - 14} = \sqrt{2 + 18 - 14} = \sqrt{6}$.The second circle is $S_{1}: (x-2 \sqrt{2})^{2}+(y-2 \sqrt{2})^{2}=r^{2}$,which has center $O(2 \sqrt{2}, 2 \sqrt{2})$.$A$ diameter of the first circle is a chord of the second circle. Let this chord be $PQ$. The distance from the center $O$ of the second circle to the center $C$ of the first circle is $d = |OC| = \sqrt{(2\sqrt{2}-\sqrt{2})^{2} + (2\sqrt{2}-3\sqrt{2})^{2}} = \sqrt{(\sqrt{2})^{2} + (-\sqrt{2})^{2}} = \sqrt{2+2} = 2$.In the right-angled triangle formed by the radius of the second circle $(r)$,the distance from the center $O$ to the chord $(d=2)$,and the radius of the first circle $(r_{1}=\sqrt{6})$,we have $r^{2} = d^{2} + r_{1}^{2}$.$r^{2} = 2^{2} + (\sqrt{6})^{2} = 4 + 6 = 10$.

If one of the diameters of the circle $x^{2}+y^{2}-2 \sqrt{2} x-6 \sqrt{2} y+14=0$ is a chord of the circle $(x-2 \sqrt{2})^{2}+(y-2 \sqrt{2})^{2}=r^{2}$,then the value of $r^{2}$ is equal to

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