If the circle `x^2+y^2 +4x+22y + c = 0` bisects the circumference of the circle `x^2 + y^2-2x + 8y-d = 0`,then `(c+d)`is equal to

Options

A. 40
B. 50
C. 60
D. 70

Answer

To solve the problem, we need to find the value of \( c + d \) given that the circle \( x^2 + y^2 + 4x + 22y + c = 0 \) bisects the circumference of the circle \( x^2 + y^2 - 2x + 8y - d = 0 \).

### Step-by-Step Solution:

1. **Write the equations of the circles**:
- Circle 1: \( x^2 + y^2 + 4x + 22y + c = 0 \)
- Circle 2: \( x^2 + y^2 - 2x + 8y - d = 0 \)

2. **Use the condition for bisection**:
When one circle bisects the other, they have a common chord. The equation of the common chord can be derived from the two circle equations:
\[ s_1 - s_2 = 0 \]
where \( s_1 \) is the equation of the first circle and \( s_2 \) is the equation of the second circle.

3. **Subtract the equations**:
\[ (x^2 + y^2 + 4x + 22y + c) - (x^2 + y^2 - 2x + 8y - d) = 0 \]
Simplifying this gives:
\[ 4x + 22y + c + 2x - 8y + d = 0 \]
\[ 6x + 14y + (c + d) = 0 \]

4. **Find the center of the second circle**:
The center of the second circle \( x^2 + y^2 - 2x + 8y - d = 0 \) can be found by comparing it to the standard form \( x^2 + y^2 + 2gx + 2fy + c = 0 \).
- Here, \( 2g = -2 \) and \( 2f = 8 \), so:
- \( g = -1 \)
- \( f = 4 \)
- Therefore, the center of the second circle is \( (1, -4) \).

5. **Substitute the center into the common chord equation**:
The common chord \( 6x + 14y + (c + d) = 0 \) must pass through the center \( (1, -4) \):
\[ 6(1) + 14(-4) + (c + d) = 0 \]
Simplifying this gives:
\[ 6 - 56 + (c + d) = 0 \]
\[ c + d - 50 = 0 \]
\[ c + d = 50 \]

### Final Answer:
Thus, the value of \( c + d \) is \( \boxed{50} \).