The locus of the centre of the circle which cuts the circles `x^(2)+y^(2)+2g_(1)x+2f_(1)y+c_(1)=0` and `x^(2)+y^(2)2g_(2)x+2f_(2)y+c_(2)=0` orthogonally is

Options

A. an ellipse
B. the radical axis of the giverr circles
C. a conic
D. another circle

Answer

To find the locus of the center of the circle that cuts the given circles orthogonally, we can follow these steps:

### Step 1: Understand the given circles
The equations of the circles are:
1. \( S_1: x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0 \)
2. \( S_2: x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0 \)

### Step 2: Identify the center of the circles
The center of a general circle given by the equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \) is at the point \( (-g, -f) \).

For the circles \( S_1 \) and \( S_2 \):
- Center of \( S_1 \) is \( (-g_1, -f_1) \)
- Center of \( S_2 \) is \( (-g_2, -f_2) \)

### Step 3: Use the orthogonality condition
For two circles to be orthogonal, the following condition must hold:
\[ 2(gg_1 + ff_1) = c + c_1 \]
For the second circle:
\[ 2(gg_2 + ff_2) = c + c_2 \]

### Step 4: Set up the equations
From the orthogonality conditions, we have:
1. \( 2(gg_1 + ff_1) = c + c_1 \) (Equation 1)
2. \( 2(gg_2 + ff_2) = c + c_2 \) (Equation 2)

### Step 5: Subtract the two equations
Subtract Equation 2 from Equation 1:
\[ 2(gg_1 - gg_2 + ff_1 - ff_2) = c_1 - c_2 \]
This simplifies to:
\[ 2(g_1 - g_2)x + 2(f_1 - f_2)y = c_1 - c_2 \]

### Step 6: Rearranging the equation
Rearranging gives us:
\[ (g_1 - g_2)x + (f_1 - f_2)y = \frac{c_1 - c_2}{2} \]

### Step 7: Identify the locus
The equation derived represents the radical axis of the two circles. The locus of the center of the circle that cuts both circles orthogonally is given by the equation of the radical axis.

### Final Answer
The locus of the center of the circle which cuts the given circles orthogonally is:
\[ (g_1 - g_2)x + (f_1 - f_2)y = \frac{c_1 - c_2}{2} \]

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