If the cirlce `(x-a)^(2)+y^(2)=25` intersects the circle `x^(2)+(y-b)^(2)=16` in such way that the legnth of the common chord is 8 units, then the vlaue of `a^(2)+b^(2)` is

Answer

To solve the problem, we need to find the value of \( a^2 + b^2 \) given the conditions of the two circles and the length of their common chord.

### Step-by-Step Solution:

1. **Identify the centers and radii of the circles:**
- The first circle is given by the equation \( (x - a)^2 + y^2 = 25 \).
- Center: \( C_1(a, 0) \)
- Radius: \( r_1 = 5 \) (since \( \sqrt{25} = 5 \))
- The second circle is given by the equation \( x^2 + (y - b)^2 = 16 \).
- Center: \( C_2(0, b) \)
- Radius: \( r_2 = 4 \) (since \( \sqrt{16} = 4 \))

2. **Understand the condition of the common chord:**
- The length of the common chord is given as \( 8 \) units.
- Since the radius of the second circle is \( 4 \), the common chord can be considered as the diameter of the second circle.

3. **Use the property of the common chord:**
- The distance from the center of one circle to the chord is perpendicular to the chord and divides it into two equal parts.
- Thus, each segment of the common chord is \( 4 \) units long.

4. **Set up the right triangle:**
- Let \( O \) be the center of the second circle \( C_2(0, b) \).
- Let \( A \) be the point where the perpendicular from \( C_1(a, 0) \) meets the chord.
- The triangle \( OAB \) is a right triangle where:
- \( OA = 4 \) (half of the common chord)
- \( OB = 5 \) (the radius of the first circle)
- \( AB = d \) (the distance between the centers \( C_1 \) and \( C_2 \))

5. **Apply the Pythagorean theorem:**
- According to the Pythagorean theorem:
\[ OB^2 = OA^2 + AB^2 \]
- Plugging in the values:
\[ 5^2 = 4^2 + d^2 \]
\[ 25 = 16 + d^2 \]
\[ d^2 = 25 - 16 = 9 \]
\[ d = 3 \]

6. **Calculate the distance \( d \) between the centers:**
- The distance \( d \) between the centers \( C_1(a, 0) \) and \( C_2(0, b) \) can be calculated using the distance formula:
\[ d = \sqrt{(a - 0)^2 + (0 - b)^2} = \sqrt{a^2 + b^2} \]
- Since we found \( d^2 = 9 \):
\[ \sqrt{a^2 + b^2} = 3 \]
\[ a^2 + b^2 = 9 \]

### Final Answer:
Thus, the value of \( a^2 + b^2 \) is \( \boxed{9} \).