The radical centre of the circles `x^2+y^2=9, x^2+y^2-2x-2y-5=0 , x^2+y^2+4x+6y-19=0` is

Options

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

Answer

To find the radical center of the given circles, we will follow these steps:

### Step 1: Write the equations of the circles in standard form
The equations of the circles are:
1. \( C_1: x^2 + y^2 = 9 \)
2. \( C_2: x^2 + y^2 - 2x - 2y - 5 = 0 \)
3. \( C_3: x^2 + y^2 + 4x + 6y - 19 = 0 \)

We can rewrite \( C_2 \) and \( C_3 \) in a more standard form:
- For \( C_2 \):
\[ x^2 + y^2 - 2x - 2y = 5 \implies (x-1)^2 + (y-1)^2 = 7 \]
- For \( C_3 \):
\[ x^2 + y^2 + 4x + 6y = 19 \implies (x+2)^2 + (y+3)^2 = 30 \]

### Step 2: Set up the equations for the radical center
The radical center can be found by solving the equations derived from the differences of the circles. We will set up two equations:
1. \( C_1 - C_2 = 0 \)
2. \( C_2 - C_3 = 0 \)

#### Equation 1: \( C_1 - C_2 = 0 \)
\[ (x^2 + y^2 - 9) - (x^2 + y^2 - 2x - 2y - 5) = 0 \]
This simplifies to:
\[ -9 + 2x + 2y + 5 = 0 \implies 2x + 2y - 4 = 0 \implies x + y = 2 \quad \text{(Equation 1)} \]

#### Equation 2: \( C_2 - C_3 = 0 \)
\[ (x^2 + y^2 - 2x - 2y - 5) - (x^2 + y^2 + 4x + 6y - 19) = 0 \]
This simplifies to:
\[ -2x - 2y - 5 - 4x - 6y + 19 = 0 \implies -6x - 8y + 14 = 0 \implies 3x + 4y = 7 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have two equations:
1. \( x + y = 2 \)
2. \( 3x + 4y = 7 \)

From Equation 1, we can express \( y \) in terms of \( x \):
\[ y = 2 - x \]

Substituting this into Equation 2:
\[ 3x + 4(2 - x) = 7 \]
\[ 3x + 8 - 4x = 7 \implies -x + 8 = 7 \implies -x = -1 \implies x = 1 \]

Now substituting \( x = 1 \) back into Equation 1:
\[ 1 + y = 2 \implies y = 1 \]

### Conclusion
The coordinates of the radical center are \( (1, 1) \).

### Final Answer
The radical center of the circles is \( (1, 1) \).

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