If the circle `x^(2) + y^(2) + 4x + 2y + c = 0` bisects the cirucumference of the cirlce `x^(2) + y^(2) -2x -8y -d = 0` then c + d =

Options

A. 60
B. `-46`
C. `40`
D. `56`

Explanation

To solve the problem, we need to find the values of \(c\) and \(d\) such that the circle given by the equation \(x^2 + y^2 + 4x + 2y + c = 0\) bisects the circumference of the circle given by \(x^2 + y^2 - 2x - 8y - d = 0\).

### Step-by-Step Solution:

1. **Identify the equations of the circles:**
- The first circle is given by:
\[ S_1: x^2 + y^2 + 4x + 2y + c = 0 \]
- The second circle is given by:
\[ S_2: x^2 + y^2 - 2x - 8y - d = 0 \]

2. **Rewrite the equations in standard form:**
- For the first circle \(S_1\):
\[ S_1: x^2 + y^2 + 4x + 2y + c = 0 \implies (x^2 + 4x) + (y^2 + 2y) + c = 0 \]
Completing the square:
\[ (x + 2)^2 - 4 + (y + 1)^2 - 1 + c = 0 \implies (x + 2)^2 + (y + 1)^2 + (c - 5) = 0 \]
Thus, the center is \((-2, -1)\) and the radius is \(\sqrt{5 - c}\).

- For the second circle \(S_2\):
\[ S_2: x^2 + y^2 - 2x - 8y - d = 0 \implies (x^2 - 2x) + (y^2 - 8y) - d = 0 \]
Completing the square:
\[ (x - 1)^2 - 1 + (y - 4)^2 - 16 - d = 0 \implies (x - 1)^2 + (y - 4)^2 - (d + 17) = 0 \]
Thus, the center is \((1, 4)\) and the radius is \(\sqrt{d + 17}\).

3. **Find the equation of the common chord:**
- The common chord of the two circles can be found using the equation \(S_1 - S_2 = 0\):
\[ (x^2 + y^2 + 4x + 2y + c) - (x^2 + y^2 - 2x - 8y - d) = 0 \]
Simplifying this gives:
\[ 4x + 2y + c + 2x + 8y + d = 0 \implies 6x + 10y + (c + d) = 0 \]

4. **Determine the condition for the common chord to bisect the circumference:**
- The common chord must pass through the center of the second circle \((1, 4)\). Therefore, substituting \(x = 1\) and \(y = 4\) into the equation of the common chord:
\[ 6(1) + 10(4) + (c + d) = 0 \]
This simplifies to:
\[ 6 + 40 + (c + d) = 0 \implies c + d + 46 = 0 \implies c + d = -46 \]

### Final Answer:
\[ c + d = -46 \]