The radical centre of the circles `x^2+y^2-x+3y-3=0 , x^2+y^2-2x+2y-2=0, x^2+y^2+2x+3y-9=0`

Options

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

Answer

To find the radical center of the circles given by the equations \(x^2+y^2-x+3y-3=0\), \(x^2+y^2-2x+2y-2=0\), and \(x^2+y^2+2x+3y-9=0\), we will follow these steps:

### Step 1: Rewrite the circle equations
We start by rewriting the equations of the circles in the form \(S = 0\).

1. Circle 1:
\[ S_1 = x^2 + y^2 - x + 3y - 3 = 0 \]

2. Circle 2:
\[ S_2 = x^2 + y^2 - 2x + 2y - 2 = 0 \]

3. Circle 3:
\[ S_3 = x^2 + y^2 + 2x + 3y - 9 = 0 \]

### Step 2: Find the radical axis of circles 1 and 2
The radical axis of two circles given by \(S_1\) and \(S_2\) is found by calculating \(S_1 - S_2 = 0\).

\[ S_1 - S_2 = (x^2 + y^2 - x + 3y - 3) - (x^2 + y^2 - 2x + 2y - 2) = 0 \]

Simplifying this gives:
\[ -x + 2x + 3y - 2y - 3 + 2 = 0 \implies x + y - 1 = 0 \implies x + y = 5 \]

### Step 3: Find the radical axis of circles 1 and 3
Next, we find the radical axis of circles 1 and 3 by calculating \(S_1 - S_3 = 0\).

\[ S_1 - S_3 = (x^2 + y^2 - x + 3y - 3) - (x^2 + y^2 + 2x + 3y - 9) = 0 \]

This simplifies to:
\[ -x - 2x + 3y - 3y - 3 + 9 = 0 \implies -3x + 6 = 0 \implies x = 2 \]

### Step 4: Solve the system of equations
Now we have two equations:
1. \(x + y = 5\) (from Step 2)
2. \(x = 2\) (from Step 3)

Substituting \(x = 2\) into the first equation:
\[ 2 + y = 5 \implies y = 3 \]

### Step 5: Conclusion
Thus, the coordinates of the radical center are:
\[ \text{Radical Center} = (2, 3) \]