If the circle `x^2 + y^2 + 6x -2y + k = 0` bisects the circumference of the circle `x^2 + y^2 + 2x- 6y-15=0` , then

Options

A. 21
B. -21
C. 23
D. -23

Explanation

To solve the problem, we need to find the value of \( k \) such that the circle given by the equation \( x^2 + y^2 + 6x - 2y + k = 0 \) bisects the circumference of the circle given by the equation \( x^2 + y^2 + 2x - 6y - 15 = 0 \).

### Step 1: Rewrite the equations of the circles
First, we rewrite both circle equations in standard form.

For the first circle:
\[ x^2 + y^2 + 6x - 2y + k = 0 \]
Completing the square for \( x \) and \( y \):
\[ (x^2 + 6x) + (y^2 - 2y) + k = 0 \]
\[ (x + 3)^2 - 9 + (y - 1)^2 - 1 + k = 0 \]
\[ (x + 3)^2 + (y - 1)^2 + (k - 10) = 0 \]
This gives us the center \( (-3, 1) \) and radius \( \sqrt{10 - k} \).

For the second circle:
\[ x^2 + y^2 + 2x - 6y - 15 = 0 \]
Completing the square:
\[ (x^2 + 2x) + (y^2 - 6y) - 15 = 0 \]
\[ (x + 1)^2 - 1 + (y - 3)^2 - 9 - 15 = 0 \]
\[ (x + 1)^2 + (y - 3)^2 - 25 = 0 \]
This gives us the center \( (-1, 3) \) and radius \( 5 \).

### Step 2: Find the equation of the common chord
The common chord of the two circles can be found by subtracting the second circle's equation from the first:
\[ (x^2 + y^2 + 6x - 2y + k) - (x^2 + y^2 + 2x - 6y - 15) = 0 \]
This simplifies to:
\[ (6x - 2y + k) - (2x - 6y - 15) = 0 \]
\[ 4x + 4y + k + 15 = 0 \]

### Step 3: Find the midpoint of the line segment joining the centers
The centers of the circles are \( C_1(-3, 1) \) and \( C_2(-1, 3) \). The midpoint \( M \) of the line segment joining these centers is:
\[ M = \left( \frac{-3 + (-1)}{2}, \frac{1 + 3}{2} \right) = \left( \frac{-4}{2}, \frac{4}{2} \right) = (-2, 2) \]

### Step 4: Substitute the midpoint into the common chord equation
Now we substitute the coordinates of the midpoint \( (-2, 2) \) into the equation of the common chord:
\[ 4(-2) + 4(2) + k + 15 = 0 \]
\[ -8 + 8 + k + 15 = 0 \]
\[ k + 15 = 0 \]
\[ k = -15 \]

### Final Answer
The value of \( k \) such that the first circle bisects the circumference of the second circle is:
\[ \boxed{-15} \]