Equation of the circle cutting orthogonally the three circles `x^(2)+y^(2)-2x+3y-7=0, x^(2)+y^(2)+5x-5y+9=0 and x^(2)+y^(2)+7x-9y+29=0` is

Options

A. `x^(2)+y^(2)-16x-18y-4=0`
B. `x^(2)+y^(2) -7x+11y+6=0`
C. `x^(2)+y^(2)+2x-8y+9=0`
D. none of these

Answer

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

### Step 1: Identify the given circles
The equations of the three circles are:
1. \( S_1: x^2 + y^2 - 2x + 3y - 7 = 0 \)
2. \( S_2: x^2 + y^2 + 5x - 5y + 9 = 0 \)
3. \( S_3: x^2 + y^2 + 7x - 9y + 29 = 0 \)

### Step 2: Find the radical axes of the circles
The radical axis of two circles \( S_1 \) and \( S_2 \) is given by \( S_1 - S_2 = 0 \).

Calculating \( S_1 - S_2 \):
\[ S_1 - S_2 = (x^2 + y^2 - 2x + 3y - 7) - (x^2 + y^2 + 5x - 5y + 9) = -7x + 8y - 16 = 0 \]
This simplifies to:
\[ -7x + 8y - 16 = 0 \quad \text{(Equation 1)} \]

Next, we find the radical axis of \( S_2 \) and \( S_3 \):
\[ S_2 - S_3 = (x^2 + y^2 + 5x - 5y + 9) - (x^2 + y^2 + 7x - 9y + 29) = -2x + 4y - 20 = 0 \]
This simplifies to:
\[ -2x + 4y - 20 = 0 \quad \text{or} \quad x - 2y + 10 = 0 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we solve the two equations obtained from the radical axes:
1. \( -7x + 8y - 16 = 0 \)
2. \( x - 2y + 10 = 0 \)

From Equation 2, we can express \( x \) in terms of \( y \):
\[ x = 2y - 10 \]

Substituting this into Equation 1:
\[ -7(2y - 10) + 8y - 16 = 0 \]
Expanding and simplifying:
\[ -14y + 70 + 8y - 16 = 0 \]
\[ -6y + 54 = 0 \]
\[ y = 9 \]

Now substituting \( y = 9 \) back into the expression for \( x \):
\[ x = 2(9) - 10 = 8 \]

Thus, the center of the circle is \( (8, 9) \).

### Step 4: Find the radius of the circle
To find the radius, we calculate the length of the tangent from the center \( (8, 9) \) to any of the circles. We will use circle \( S_1 \):
\[ S_1: x^2 + y^2 - 2x + 3y - 7 = 0 \]
The radius \( r \) is given by:
\[ r = \sqrt{(h - a)^2 + (k - b)^2 - c} \]
where \( (h, k) \) is the center of the new circle, \( (a, b) \) is the center of the circle \( S_1 \), and \( c \) is the radius squared of circle \( S_1 \).

The center of \( S_1 \) can be found from its equation:
\[ x^2 + y^2 - 2x + 3y - 7 = 0 \implies (x - 1)^2 + (y + \frac{3}{2})^2 = \frac{49}{4} \]
So, the center is \( (1, -\frac{3}{2}) \) and the radius is \( r = \frac{7}{2} \).

Now calculating the length of the tangent:
\[ r = \sqrt{(8 - 1)^2 + (9 + \frac{3}{2})^2 - \left(\frac{7}{2}\right)^2} \]
Calculating:
\[ = \sqrt{(7)^2 + (10.5)^2 - \left(\frac{7}{2}\right)^2} \]
\[ = \sqrt{49 + 110.25 - 12.25} = \sqrt{147} = \sqrt{149} \]

### Step 5: Write the equation of the circle
The general equation of a circle with center \( (h, k) \) and radius \( r \) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \( h = 8, k = 9, r = \sqrt{149} \):
\[ (x - 8)^2 + (y - 9)^2 = 149 \]
Expanding this:
\[ x^2 - 16x + 64 + y^2 - 18y + 81 = 149 \]
Combining terms:
\[ x^2 + y^2 - 16x - 18y + 145 - 149 = 0 \]
Thus, the final equation of the circle is:
\[ x^2 + y^2 - 16x - 18y - 4 = 0 \]

### Final Answer
The equation of the circle is:
\[ \boxed{x^2 + y^2 - 16x - 18y - 4 = 0} \]