1. Find the first circle's center and radius

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

Compare with the standard form: $x^2+y^2-2gx-2fy+c=0$ So, $g=\sqrt{2},\quad f=3\sqrt{2},\quad c=14$

Hence the center is $C_1=(\sqrt{2},3\sqrt{2})$

Radius: $R_1=\sqrt{g^2+f^2-c}$ $=\sqrt{2+18-14}=\sqrt{6}$

So the first circle is centered at $C_1=(\sqrt{2},3\sqrt{2})$ with radius $\sqrt{6}$.

2. Identify the given diameter of the first circle

A diameter of the first circle can be any line segment through its center with endpoints on the circle. The question says that one of the diameters of the first circle is a chord of the second circle.

This means that this diameter, whose length is $2R_1=2\sqrt{6},$ is a chord of the second circle.

Also, since every diameter of the first circle passes through $C_1$, the required chord of the second circle passes through $C_1$.

For a fixed point inside a circle, the longest chord through that point is the chord perpendicular to the radius through that point, and its length depends on the distance of the point from the center. But here the statement says that a diameter of the first circle is a chord of the second circle. So the line through $C_1$ containing this diameter must intersect the second circle at exactly the endpoints of that diameter.

Thus the point $C_1$ must lie on the second circle in such a way that the chord through it has half-length $\sqrt{6}$.

A more direct approach: since the diameter of the first circle has endpoints on the first circle and also lies as a chord of the second circle, the distance from the center of the second circle to the line of this chord determines the chord length.

3. Find the center of the second circle

Given second circle: $(x-2\sqrt{2})^2+(y-2\sqrt{2})^2=r^2$

Its center is $C_2=(2\sqrt{2},2\sqrt{2})$

Now find the distance between the centers: $C_1C_2=\sqrt{(2\sqrt{2}-\sqrt{2})^2+(2\sqrt{2}-3\sqrt{2})^2}$ $=\sqrt{(\sqrt{2})^2+(-\sqrt{2})^2}$ $=\sqrt{2+2}=2$

So the point $C_1$ is at distance $2$ from the center $C_2$.

4. Use chord-length formula in the second circle

A chord of a circle of radius $r$, at perpendicular distance $d$ from the center, has length $L=2\sqrt{r^2-d^2}$

Here, the chord is actually a diameter of the first circle, so $L=2\sqrt{6}$

Since this chord passes through $C_1$, and we can choose the diameter so that it becomes a chord of the second circle, the perpendicular distance from $C_2$ to this chord can be at most $C_1C_2=2$. For the required diameter through $C_1$, the suitable line is perpendicular to $C_1C_2$, so the distance from $C_2$ to the chord equals $d=C_1C_2=2$

Therefore, $2\sqrt{r^2-2^2}=2\sqrt{6}$

Divide by $2$: $\sqrt{r^2-4}=\sqrt{6}$

Square both sides: $r^2-4=6$ $r^2=10$

5. Final answer

$\boxed{10}$

This matches the stored correct answer.