The centre of the circle which intersects the three circles, `x^(2)+y^(2)+4x+7=0, x^(2)+y^(2)+y=0 and 2x^(2)+2y^(2)+3x+5y+9=0` orthogonally is the point

Options

A. (2, 1)
B. (-2, 1)
C. (2, -1)
D. (-2, -1)

Answer

To find the center of the circle that intersects the three given circles orthogonally, we will follow these steps:

### Step 1: Rewrite the equations of the circles in standard form
The equations of the circles are given as:
1. \( x^2 + y^2 + 4x + 7 = 0 \)
2. \( x^2 + y^2 + y = 0 \)
3. \( 2x^2 + 2y^2 + 3x + 5y + 9 = 0 \)

We can rewrite these equations in the standard form of a circle, which is \( x^2 + y^2 + 2gx + 2fy + c = 0 \).

1. For the first circle:
\[ x^2 + y^2 + 4x + 7 = 0 \implies 2g = 4 \implies g_1 = 2, \quad c_1 = 7 \]

2. For the second circle:
\[ x^2 + y^2 + y = 0 \implies 2g = 0, \quad 2f = 1 \implies g_2 = 0, \quad f_2 = \frac{1}{2}, \quad c_2 = 0 \]

3. For the third circle:
\[ 2x^2 + 2y^2 + 3x + 5y + 9 = 0 \implies x^2 + y^2 + \frac{3}{2}x + \frac{5}{2}y + \frac{9}{2} = 0 \]
\[ \implies 2g_3 = \frac{3}{2} \implies g_3 = \frac{3}{4}, \quad 2f_3 = \frac{5}{2} \implies f_3 = \frac{5}{4}, \quad c_3 = \frac{9}{2} \]

### Step 2: Set up orthogonality conditions
For two circles to intersect orthogonally, the following condition must hold:
\[ g_1g + f_1f = c + c_1 \]
where \( g, f, c \) are the coefficients of the circle we are trying to find.

1. For the first circle:
\[ 2g + 0 \cdot f = c + 7 \implies 2g = c + 7 \quad \text{(Equation 1)} \]

2. For the second circle:
\[ 0 \cdot g + \frac{1}{2} f = c + 0 \implies \frac{1}{2} f = c \quad \text{(Equation 2)} \]

3. For the third circle:
\[ \frac{3}{4} g + \frac{5}{4} f = c + \frac{9}{2} \quad \text{(Equation 3)} \]

### Step 3: Solve the equations
From Equation 2, we can express \( c \) in terms of \( f \):
\[ c = \frac{1}{2} f \]

Substituting \( c \) into Equation 1:
\[ 2g = \frac{1}{2} f + 7 \implies 4g = f + 14 \implies f = 4g - 14 \quad \text{(Equation 4)} \]

Now substitute \( f \) from Equation 4 into Equation 3:
\[ \frac{3}{4} g + \frac{5}{4}(4g - 14) = \frac{1}{2}(4g - 14) + \frac{9}{2} \]
Simplifying:
\[ \frac{3}{4} g + 5g - \frac{70}{4} = 2g - 7 + \frac{9}{2} \]
Combine like terms:
\[ \frac{3}{4} g + 5g = \frac{20}{4} g \implies \frac{20}{4} g - 2g = \frac{70}{4} - \frac{18}{4} \]
This simplifies to:
\[ \frac{20}{4} g - \frac{8}{4} g = \frac{52}{4} \implies \frac{12}{4} g = \frac{52}{4} \implies 3g = 13 \implies g = \frac{13}{3} \]

### Step 4: Find \( f \) and \( c \)
Substituting \( g \) back into Equation 4:
\[ f = 4 \cdot \frac{13}{3} - 14 = \frac{52}{3} - \frac{42}{3} = \frac{10}{3} \]

And substituting \( f \) back into Equation 2:
\[ c = \frac{1}{2} \cdot \frac{10}{3} = \frac{5}{3} \]

### Step 5: Find the center of the circle
The center of the circle is given by the coordinates \( (-g, -f) \):
\[ \text{Center} = \left(-\frac{13}{3}, -\frac{10}{3}\right) \]

### Final Answer
The center of the circle that intersects the three given circles orthogonally is:
\[ \boxed{\left(-\frac{13}{3}, -\frac{10}{3}\right)} \]