The value of k for which the circle `x ^(2) +y ^(2) - 4x + 6y + 3=0` will bisect the circumference of the circle `x ^(2) + y ^(2) + 6x - 4y + k =0` is

Options

A. 53
B. `-53`
C. 47
D. `-47`

Answer

To find the value of \( k \) for which the circle \( x^2 + y^2 - 4x + 6y + 3 = 0 \) bisects the circumference of the circle \( x^2 + y^2 + 6x - 4y + k = 0 \), we can follow these steps:

### Step 1: Rewrite the equations of the circles in standard form

1. **First Circle**:
\[ x^2 + y^2 - 4x + 6y + 3 = 0 \]
Completing the square:
\[ (x^2 - 4x) + (y^2 + 6y) + 3 = 0 \]
\[ (x - 2)^2 - 4 + (y + 3)^2 - 9 + 3 = 0 \]
\[ (x - 2)^2 + (y + 3)^2 = 10 \]
The center \( C_1 \) is \( (2, -3) \) and the radius \( r_1 = \sqrt{10} \).

2. **Second Circle**:
\[ x^2 + y^2 + 6x - 4y + k = 0 \]
Completing the square:
\[ (x^2 + 6x) + (y^2 - 4y) + k = 0 \]
\[ (x + 3)^2 - 9 + (y - 2)^2 - 4 + k = 0 \]
\[ (x + 3)^2 + (y - 2)^2 = 13 - k \]
The center \( C_2 \) is \( (-3, 2) \) and the radius \( r_2 = \sqrt{13 - k} \).

### Step 2: Find the common chord equation

The common chord of the two circles can be found using the equation:
\[ S_1 - S_2 = 0 \]
Where \( S_1 \) is the equation of the first circle and \( S_2 \) is the equation of the second circle.

Substituting the equations:
\[ (x^2 + y^2 - 4x + 6y + 3) - (x^2 + y^2 + 6x - 4y + k) = 0 \]
This simplifies to:
\[ -4x + 6y + 3 - 6x + 4y - k = 0 \]
\[ -10x + 10y + 3 - k = 0 \]
Rearranging gives us:
\[ 10y - 10x + (3 - k) = 0 \]
Or:
\[ y - x + \frac{3 - k}{10} = 0 \]

### Step 3: Check if the center of the second circle lies on the common chord

The center \( C_2 = (-3, 2) \) must satisfy the equation of the common chord:
\[ 2 - (-3) + \frac{3 - k}{10} = 0 \]
This simplifies to:
\[ 5 + \frac{3 - k}{10} = 0 \]
Multiplying through by 10:
\[ 50 + 3 - k = 0 \]
\[ k = 53 \]

### Conclusion

The value of \( k \) for which the first circle bisects the circumference of the second circle is:
\[ \boxed{53} \]