The centre of the circle orthogonal to the circles `x^2+y^2+4y+1=0 , x^2+y^2+6x+y-8=0, x^2+y^2-4x-4y+37=0` is

Options

A. (1,1)
B. (2,2)
C. (3,3)
D. (0,0)

Answer

To find the center of the circle orthogonal to the given circles, we will follow these steps:

### Step 1: Write the equations of the circles in standard form
The given equations of the circles are:
1. \( x^2 + y^2 + 4y + 1 = 0 \)
2. \( x^2 + y^2 + 6x + y - 8 = 0 \)
3. \( x^2 + y^2 - 4x - 4y + 37 = 0 \)

We need to rewrite these equations in the standard form of a circle, which is \( (x - h)^2 + (y - k)^2 = r^2 \).

#### For Circle 1:
\[ x^2 + y^2 + 4y + 1 = 0 \]
Rearranging gives:
\[ x^2 + (y^2 + 4y + 4) - 4 + 1 = 0 \]
\[ x^2 + (y + 2)^2 - 3 = 0 \]
Thus, the center is \( (0, -2) \) and radius \( r = \sqrt{3} \).

#### For Circle 2:
\[ x^2 + y^2 + 6x + y - 8 = 0 \]
Rearranging gives:
\[ (x^2 + 6x + 9) + (y^2 + y + \frac{1}{4}) - 9 - \frac{1}{4} - 8 = 0 \]
\[ (x + 3)^2 + (y + \frac{1}{2})^2 - \frac{81}{4} = 0 \]
Thus, the center is \( (-3, -\frac{1}{2}) \) and radius \( r = \frac{9}{2} \).

#### For Circle 3:
\[ x^2 + y^2 - 4x - 4y + 37 = 0 \]
Rearranging gives:
\[ (x^2 - 4x + 4) + (y^2 - 4y + 4) - 4 + 37 = 0 \]
\[ (x - 2)^2 + (y - 2)^2 - 33 = 0 \]
Thus, the center is \( (2, 2) \) and radius \( r = \sqrt{33} \).

### Step 2: Set up the equations for orthogonality
For two circles to be orthogonal, the following condition must hold:
\[ S_1 S_2 = r_1^2 + r_2^2 \]
Where \( S_1 \) and \( S_2 \) are the equations of the circles.

We will set up the equations based on the centers and radii of the circles.

### Step 3: Solve the equations
We will derive two equations from the orthogonality condition and solve for the center of the orthogonal circle.

1. From Circle 1 and Circle 2:
\[ S_1 - S_2 = 0 \]
This leads to:
\[ -6x + 3y + 9 = 0 \] (Equation 1)

2. From Circle 2 and Circle 3:
\[ S_2 - S_3 = 0 \]
This leads to:
\[ 2x + y - 9 = 0 \] (Equation 2)

### Step 4: Solve the system of equations
Now we will solve the two equations obtained:

1. From Equation 1:
\[ -6x + 3y + 9 = 0 \]
Rearranging gives:
\[ 3y = 6x - 9 \]
\[ y = 2x - 3 \]

2. Substitute \( y \) in Equation 2:
\[ 2x + (2x - 3) - 9 = 0 \]
\[ 4x - 12 = 0 \]
\[ x = 3 \]

3. Substitute \( x \) back to find \( y \):
\[ y = 2(3) - 3 = 6 - 3 = 3 \]

### Final Answer:
The center of the circle orthogonal to the given circles is \( (3, 3) \).

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