If one of the diameters of the circle x ^ 2 + y ^ 2 - 2 2 x - 6 2 y...

Given that, one of the diameter of the circle ${\left(x - \sqrt{2}\right)}^{2} + {\left(y - 3\sqrt{2}\right)}^{2} = 6$ is a chord of another circle ${\left(x - 2\sqrt{2}\right)}^{2} + {\left(y - 2\sqrt{2}\right)}^{2} = {r}^{2}$.

Let, $C$ and ${C}_{1}$ be the centres of the circles ${\left(x - \sqrt{2}\right)}^{2} + {\left(y - 3\sqrt{2}\right)}^{2} = 6$ and ${\left(x - 2\sqrt{2}\right)}^{2} + {\left(y - 2\sqrt{2}\right)}^{2} = {r}^{2}$ respectively.

So, $C\left(\sqrt{2}, 3\sqrt{2}\right) \text{ and } {C}_{1}\left(2\sqrt{2}, 2\sqrt{2}\right)$

The position of the two circles as shown in figure.

Now, $C{C}_{1} = \sqrt{{\left(2\sqrt{2} - \sqrt{2}\right)}^{2} + {\left(2\sqrt{2} - 3\sqrt{2}\right)}^{2}} = \sqrt{2 + 2} = 2$

It is clear from the above diagram, $∆ AC{C}_{1}$ is a right angled triangle.$⇒ {r}^{2} = {\left(AC\right)}^{2} + {\left(C{C}_{1}\right)}^{2} = {\left(\sqrt{6}\right)}^{2} + {2}^{2}$ $⇒ {r}^{2} = 10$