If the circle `x^2 + y^2 + 2x - 2y + 4 = 0` cuts the circle `x^2 + y^2 + 4x - 2fy +2 = 0` orthogonally, then f =

Options

A. 1
B. 2
C. `-1`
D. `-2`

Answer

To solve the problem, we need to determine the value of \( f \) such that the two given circles intersect orthogonally.

### Step-by-step Solution:

1. **Write the equations of the circles**:
The first circle is given by:
\[ x^2 + y^2 + 2x - 2y + 4 = 0 \]
The second circle is given by:
\[ x^2 + y^2 + 4x - 2fy + 2 = 0 \]

2. **Identify the coefficients**:
For the first circle, we can compare it with the general equation of a circle:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From the first circle, we have:
- \( g_1 = 1 \)
- \( f_1 = -1 \)
- \( c_1 = 4 \)

For the second circle, we compare it similarly:
\[ x^2 + y^2 + 4x - 2fy + 2 = 0 \]
From this, we have:
- \( g_2 = 2 \)
- \( f_2 = f \)
- \( c_2 = 2 \)

3. **Use the orthogonality condition**:
The condition for two circles to intersect orthogonally is given by:
\[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \]
Substituting the values we found:
\[ 2(1)(2) + 2(-1)(f) = 4 + 2 \]

4. **Simplify the equation**:
Simplifying the left side:
\[ 4 - 2f = 6 \]

5. **Solve for \( f \)**:
Rearranging the equation:
\[ -2f = 6 - 4 \]
\[ -2f = 2 \]
Dividing both sides by -2:
\[ f = -1 \]

### Final Answer:
Thus, the value of \( f \) is:
\[ \boxed{-1} \]