Equation of the circle which cuts each of the circles `x ^(2) + y ^(2) + 2gx + c =0, x ^(2) + y ^(2) + 2g _(1) x +c =0 and x ^(2) + y^(2) + 2hx + 2ky + a=0` orthogonally is

Options

A. `h (x ^(2) +y ^(2)) + (a-c) ch =0`
B. `k (x ^(2) + y ^(2)) +(a-c) y-ck =0`
C. `x ^(2) + y ^(2) + (a-c) + c=0`
D. `x ^(2) + y ^(2) + (a-c) y -x =0`

Answer

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

### Step 1: Write the general equation of the circle
Let the equation of the circle be:
\[ x^2 + y^2 + 2px + 2qy + d = 0 \]
where \( p \), \( q \), and \( d \) are constants to be determined.

### Step 2: Identify the given circles
The given circles are:
1. \( x^2 + y^2 + 2gx + c = 0 \)
2. \( x^2 + y^2 + 2g_1x + c = 0 \)
3. \( x^2 + y^2 + 2hx + 2ky + a = 0 \)

### Step 3: Use the orthogonality condition
For two circles to be orthogonal, the following condition must hold:
\[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \]
where \( g_1, g_2 \) are the coefficients of \( x \), \( f_1, f_2 \) are the coefficients of \( y \), and \( c_1, c_2 \) are the constant terms.

### Step 4: Apply the orthogonality condition for the first circle
For the first circle:
- \( g_1 = p \), \( f_1 = q \), \( c_1 = d \)
- \( g_2 = g \), \( f_2 = 0 \), \( c_2 = c \)

Applying the orthogonality condition:
\[ 2pg + 0 = d + c \quad \text{(Equation 1)} \]

### Step 5: Apply the orthogonality condition for the second circle
For the second circle:
- \( g_1 = p \), \( f_1 = q \), \( c_1 = d \)
- \( g_2 = g_1 \), \( f_2 = 0 \), \( c_2 = c \)

Applying the orthogonality condition:
\[ 2pg_1 + 0 = d + c \quad \text{(Equation 2)} \]

### Step 6: Apply the orthogonality condition for the third circle
For the third circle:
- \( g_1 = p \), \( f_1 = q \), \( c_1 = d \)
- \( g_2 = h \), \( f_2 = k \), \( c_2 = a \)

Applying the orthogonality condition:
\[ 2ph + 2qk = d + a \quad \text{(Equation 3)} \]

### Step 7: Solve the equations
From Equation 1 and Equation 2, since both equal \( d + c \):
\[ 2pg = 2pg_1 \]
This implies:
\[ p(g - g_1) = 0 \]
Thus, either \( p = 0 \) or \( g = g_1 \).

### Step 8: Substitute \( p = 0 \)
If \( p = 0 \), substituting into Equation 1 gives:
\[ 0 = d + c \implies d = -c \]

### Step 9: Substitute \( p = 0 \) and \( d = -c \) into Equation 3
Substituting into Equation 3:
\[ 0 + 2qk = -c + a \implies 2qk = a - c \implies q = \frac{a - c}{2k} \]

### Step 10: Write the final equation of the circle
Substituting \( p = 0 \) and \( d = -c \) into the general equation of the circle:
\[ x^2 + y^2 + 0 + 2qy - c = 0 \]
This simplifies to:
\[ x^2 + y^2 + (a - c)y - c = 0 \]

### Final Equation
Thus, the equation of the circle that cuts each of the given circles orthogonally is:
\[ x^2 + y^2 + \frac{(a - c)}{k}y - c = 0 \]