If one of the diameter of the circle `x^(2) +y^(2) - 2 sqrt2x- 6 sqrt2y + 14 = 0` is a chord of the circle `(x- 2 sqrt2)^(2) + (y-2 sqrt2)^(2)= r^(2)`, then the value of `r^(2)` is equal to _____

Answer

To solve the problem, we need to find the value of \( r^2 \) given the conditions of the circles. Let's break it down step by step.

### Step 1: Rewrite the first circle's equation
The equation of the first circle is given as:
\[ x^2 + y^2 - 2\sqrt{2}x - 6\sqrt{2}y + 14 = 0 \]
We can rearrange this into the standard form of a circle, which is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
To do this, we complete the square for both \( x \) and \( y \).

### Step 2: Completing the square
For \( x \):
\[ x^2 - 2\sqrt{2}x = (x - \sqrt{2})^2 - 2 \]
For \( y \):
\[ y^2 - 6\sqrt{2}y = (y - 3\sqrt{2})^2 - 18 \]
Now substituting back into the equation:
\[ (x - \sqrt{2})^2 - 2 + (y - 3\sqrt{2})^2 - 18 + 14 = 0 \]
This simplifies to:
\[ (x - \sqrt{2})^2 + (y - 3\sqrt{2})^2 - 6 = 0 \]
Thus, we have:
\[ (x - \sqrt{2})^2 + (y - 3\sqrt{2})^2 = 6 \]
From this, we can identify the center and radius:
- Center: \( (\sqrt{2}, 3\sqrt{2}) \)
- Radius: \( \sqrt{6} \)

### Step 3: Identify the second circle's equation
The second circle is given as:
\[ (x - 2\sqrt{2})^2 + (y - 2\sqrt{2})^2 = r^2 \]
The center of this circle is \( (2\sqrt{2}, 2\sqrt{2}) \).

### Step 4: Determine the distance between the centers
We need to find the distance between the centers of the two circles:
- Center of the first circle: \( (\sqrt{2}, 3\sqrt{2}) \)
- Center of the second circle: \( (2\sqrt{2}, 2\sqrt{2}) \)

Using the distance formula:
\[ d = \sqrt{(2\sqrt{2} - \sqrt{2})^2 + (2\sqrt{2} - 3\sqrt{2})^2} \]
Calculating this gives:
\[ d = \sqrt{(\sqrt{2})^2 + (-\sqrt{2})^2} = \sqrt{2 + 2} = \sqrt{4} = 2 \]

### Step 5: Apply the Pythagorean theorem
Since one of the diameters of the first circle is a chord of the second circle, we can apply the Pythagorean theorem:
\[ AC^2 = AB^2 + BC^2 \]
Where:
- \( AB \) is the radius of the first circle: \( \sqrt{6} \)
- \( BC \) is the distance from the center of the second circle to the midpoint of the diameter of the first circle, which we found to be \( 2 \).

Thus:
\[ AC^2 = r^2 \]
Substituting the known values:
\[ r^2 = (\sqrt{6})^2 + (2)^2 \]
Calculating this gives:
\[ r^2 = 6 + 4 = 10 \]

### Final Answer
The value of \( r^2 \) is \( \boxed{10} \).