Find the equation of the circle which cuts each of the following circles orthogonally.
`x^2+y^2+2x+17y+4=0`
`x^2+y^2+7x+6y+11=0`
`x^2+y^2-x+22y+3=0`

Answer

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

### Step 1: Write the equations of the given circles in standard form

The given circles are:
1. \( x^2 + y^2 + 2x + 17y + 4 = 0 \)
2. \( x^2 + y^2 + 7x + 6y + 11 = 0 \)
3. \( x^2 + y^2 - x + 22y + 3 = 0 \)

### Step 2: Identify the centers and radii of the given circles

To find the centers and radii, we can rewrite each equation in the standard form \((x - h)^2 + (y - k)^2 = r^2\).

1. **Circle 1**:
\[ x^2 + 2x + y^2 + 17y + 4 = 0 \implies (x + 1)^2 + (y + \frac{17}{2})^2 = \frac{57}{4} \]
Center: \((-1, -\frac{17}{2})\), Radius: \(\frac{\sqrt{57}}{2}\)

2. **Circle 2**:
\[ x^2 + 7x + y^2 + 6y + 11 = 0 \implies (x + \frac{7}{2})^2 + (y + 3)^2 = \frac{25}{4} \]
Center: \((- \frac{7}{2}, -3)\), Radius: \(\frac{5}{2}\)

3. **Circle 3**:
\[ x^2 - x + y^2 + 22y + 3 = 0 \implies (x - \frac{1}{2})^2 + (y + 11)^2 = \frac{121}{4} \]
Center: \((\frac{1}{2}, -11)\), Radius: \(\frac{11}{2}\)

### Step 3: Find the radical axis of the circles

To find the radical axis, we subtract the equations of the circles pairwise.

1. **Subtract Circle 1 from Circle 2**:
\[ (x^2 + y^2 + 7x + 6y + 11) - (x^2 + y^2 + 2x + 17y + 4) = 0 \]
This simplifies to:
\[ 5x - 11y + 7 = 0 \quad \text{(Equation 4)} \]

2. **Subtract Circle 2 from Circle 3**:
\[ (x^2 + y^2 - x + 22y + 3) - (x^2 + y^2 + 7x + 6y + 11) = 0 \]
This simplifies to:
\[ -8x + 16y - 8 = 0 \quad \text{(Equation 5)} \]

### Step 4: Solve the system of equations (Equations 4 and 5)

From Equation 4:
\[ 5x - 11y + 7 = 0 \implies 5x = 11y - 7 \implies x = \frac{11y - 7}{5} \]

Substituting \(x\) into Equation 5:
\[ -8\left(\frac{11y - 7}{5}\right) + 16y - 8 = 0 \]
Multiply through by 5 to eliminate the fraction:
\[ -8(11y - 7) + 80y - 40 = 0 \]
\[ -88y + 56 + 80y - 40 = 0 \implies -8y + 16 = 0 \implies y = 2 \]

Substituting \(y = 2\) back to find \(x\):
\[ x = \frac{11(2) - 7}{5} = \frac{22 - 7}{5} = \frac{15}{5} = 3 \]

Thus, the center of the new circle is \((3, 2)\).

### Step 5: Find the radius of the new circle

To find the radius, we can substitute the center \((3, 2)\) into any of the original circle equations. Let's use Circle 1:
\[ 3^2 + 2^2 + 2(3) + 17(2) + 4 = 0 \]
Calculating:
\[ 9 + 4 + 6 + 34 + 4 = 57 \]
Thus, the radius \(r\) is:
\[ r = \sqrt{57} \]

### Step 6: Write the equation of the new circle

The equation of the circle with center \((3, 2)\) and radius \(\sqrt{57}\) is:
\[ (x - 3)^2 + (y - 2)^2 = 57 \]

Expanding this gives:
\[ x^2 - 6x + 9 + y^2 - 4y + 4 - 57 = 0 \implies x^2 + y^2 - 6x - 4y - 44 = 0 \]

### Final Answer
The equation of the required circle is:
\[ x^2 + y^2 - 6x - 4y - 44 = 0 \]