The equation of the circle which cuts orthogonally the three circles `x^2+y^2+4x+2y+1=0, 2x^2+2y^2+8x+6y-3=0, x^2+y^2+6x-2y-3=0` is

Options

A. `x^2+y^2-6x-4y-44=0`
B. `x^2+y^2-6x+6=0`
C. `x^2+y^2-14x-5y-34=0`
D. `x^2+y^2-5x-14y-34=0`

Answer

To solve the problem of finding the equation of the circle that cuts orthogonally the three given circles, we will follow these steps:

### Step 1: Write down the equations of the circles
We have three circles given by their equations:
1. \( S_1: x^2 + y^2 + 4x + 2y + 1 = 0 \)
2. \( S_2: 2x^2 + 2y^2 + 8x + 6y - 3 = 0 \)
3. \( S_3: x^2 + y^2 + 6x - 2y - 3 = 0 \)

### Step 2: Normalize the second circle
To make the coefficients of \( x^2 \) and \( y^2 \) equal to 1, we divide the entire equation of \( S_2 \) by 2:
\[ S_2: x^2 + y^2 + 4x + 3y - \frac{3}{2} = 0 \]

### Step 3: Use the radical axis condition
The radical axis condition states that for two circles \( S_1 \) and \( S_2 \), the equation of the radical axis is given by:
\[ S_1 - S_2 = 0 \]
Substituting the equations of \( S_1 \) and \( S_2 \):
\[ (x^2 + y^2 + 4x + 2y + 1) - (x^2 + y^2 + 4x + 3y - \frac{3}{2}) = 0 \]
This simplifies to:
\[ 2y + 1 - 3y + \frac{3}{2} = 0 \]
Combining like terms gives:
\[ -y + \frac{5}{2} = 0 \implies y = \frac{5}{2} \]

### Step 4: Find \( x \) using \( S_1 \) and \( S_3 \)
Now we will use the radical axis condition between \( S_1 \) and \( S_3 \):
\[ S_1 - S_3 = 0 \]
Substituting the equations:
\[ (x^2 + y^2 + 4x + 2y + 1) - (x^2 + y^2 + 6x - 2y - 3) = 0 \]
This simplifies to:
\[ 4x + 2y + 1 - 6x + 2y + 3 = 0 \]
Combining like terms gives:
\[ -2x + 4y + 4 = 0 \implies x - 2y - 2 = 0 \implies x = 2y + 2 \]
Substituting \( y = \frac{5}{2} \):
\[ x = 2 \cdot \frac{5}{2} + 2 = 5 + 2 = 7 \]

### Step 5: Determine the center of the circle
The center of the circle is \( (7, \frac{5}{2}) \).

### Step 6: Find the radius of the circle
Using the first circle \( S_1 \) to find the radius:
\[ r = \sqrt{x^2 + y^2 + 4x + 2y + 1} \]
Substituting \( x = 7 \) and \( y = \frac{5}{2} \):
\[ r = \sqrt{7^2 + \left(\frac{5}{2}\right)^2 + 4 \cdot 7 + 2 \cdot \frac{5}{2} + 1} \]
Calculating each term:
\[ = \sqrt{49 + \frac{25}{4} + 28 + 5 + 1} = \sqrt{49 + 28 + 5 + 1 + \frac{25}{4}} = \sqrt{78 + \frac{25}{4}} = \sqrt{\frac{312 + 25}{4}} = \sqrt{\frac{337}{4}} = \frac{\sqrt{337}}{2} \]

### Step 7: Write the equation of the circle
The equation of the circle in standard form is:
\[ (x - 7)^2 + \left(y - \frac{5}{2}\right)^2 = \left(\frac{\sqrt{337}}{2}\right)^2 \]
Expanding this gives:
\[ (x - 7)^2 + \left(y - \frac{5}{2}\right)^2 = \frac{337}{4} \]
This can be expanded further to:
\[ x^2 - 14x + 49 + y^2 - 5y + \frac{25}{4} = \frac{337}{4} \]
Rearranging gives:
\[ x^2 + y^2 - 14x - 5y + 49 + \frac{25}{4} - \frac{337}{4} = 0 \]
Combining constants:
\[ x^2 + y^2 - 14x - 5y - \frac{263}{4} = 0 \]
Multiplying through by 4 to eliminate the fraction:
\[ 4x^2 + 4y^2 - 56x - 20y - 263 = 0 \]

### Final Answer
The required equation of the circle is:
\[ 4x^2 + 4y^2 - 56x - 20y - 263 = 0 \]