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

Given circle,

$$x^2+y^2-2\sqrt2x-6\sqrt2y+14=0$$

Comparing with

$$x^2+y^2+2gx+2fy+c=0$$

we get

$$g=-\sqrt2,\qquad f=-3\sqrt2$$

Hence, the centre is

$$C_1=(\sqrt2,3\sqrt2)$$

and radius

$$R=\sqrt{g^2+f^2-c}$$

$$=\sqrt{2+18-14}=\sqrt6$$

The second circle is

$$ (x-2\sqrt2)^2+(y-2\sqrt2)^2=r^2 $$

whose centre is

$$C_2=(2\sqrt2,2\sqrt2)$$

Now, one diameter of the first circle is a chord of the second circle.

Hence, the length of the chord of the second circle equals the diameter of the first circle.

So,

$$\text{Chord length}=2\sqrt6$$

Distance of the chord from centre $$C_2$$ equals the perpendicular distance from $$C_2$$ to the diameter line.

Since every diameter passes through $$C_1$$, the required diameter is the line through $$C_1$$ perpendicular to $$C_1C_2$$.

Thus,

$$d=|C_1C_2|$$

Now,

$$|C_1C_2| = \sqrt{(\sqrt2)^2+(-\sqrt2)^2} = 2$$

Using chord formula,

$$2\sqrt{r^2-d^2}=2\sqrt6$$

$$r^2-d^2=6$$

$$r^2-4=6$$

$$r^2=10$$

Hence,

$$\boxed{10}$$