If the circle `x^2+y^2-6x-4y+9=0` bisects the circumference of the circle `x^2+y^2-8x-6y+a=0` , then the value of a is ____

Answer

To solve the problem, we need to find the value of \( a \) such that the circle given by the equation \( x^2 + y^2 - 8x - 6y + a = 0 \) is bisected by the circle given by the equation \( x^2 + y^2 - 6x - 4y + 9 = 0 \).

### Step-by-Step Solution:

1. **Identify the center and radius of the first circle**:
The equation of the first circle is:
\[ x^2 + y^2 - 6x - 4y + 9 = 0 \]
We can rewrite it in standard form by completing the square:
\[ (x^2 - 6x) + (y^2 - 4y) = -9 \]
Completing the square:
\[ (x - 3)^2 - 9 + (y - 2)^2 - 4 = -9 \]
\[ (x - 3)^2 + (y - 2)^2 = 4 \]
Hence, the center \( C_1 \) is \( (3, 2) \) and the radius \( R_1 \) is \( 2 \).

2. **Identify the center and radius of the second circle**:
The equation of the second circle is:
\[ x^2 + y^2 - 8x - 6y + a = 0 \]
Similarly, we rewrite it in standard form:
\[ (x^2 - 8x) + (y^2 - 6y) = -a \]
Completing the square:
\[ (x - 4)^2 - 16 + (y - 3)^2 - 9 = -a \]
\[ (x - 4)^2 + (y - 3)^2 = a + 25 \]
Thus, the center \( C_2 \) is \( (4, 3) \) and the radius \( R_2 \) is \( \sqrt{a + 25} \).

3. **Condition for bisecting the circumference**:
For the first circle to bisect the second circle, the line joining the centers \( C_1 \) and \( C_2 \) must be a diameter of the second circle. The midpoint of the line segment connecting the centers is:
\[ \left( \frac{3 + 4}{2}, \frac{2 + 3}{2} \right) = \left( \frac{7}{2}, \frac{5}{2} \right) \]
The distance between the centers \( C_1 \) and \( C_2 \) is:
\[ d = \sqrt{(4 - 3)^2 + (3 - 2)^2} = \sqrt{1 + 1} = \sqrt{2} \]
For the first circle to bisect the second, the distance \( d \) must equal the radius of the second circle \( R_2 \):
\[ \sqrt{2} = \sqrt{a + 25} \]

4. **Square both sides to eliminate the square root**:
\[ 2 = a + 25 \]
Rearranging gives:
\[ a = 2 - 25 = -23 \]

### Final Answer:
The value of \( a \) is \( -23 \).