The radical centre of `x^(2)+y^(2)-2x-2y+1=0, x^(2)+y^(2)-4x+6y-2=0 and x^(2)+y^(2)+3x-2y+4=0` is

Options

A. `((3)/(5),(9)/(40))`
B. `(-(3)/(5),-(9)/(40))`
C. `(-(3)/(5),(9)/(40))`
D. `((3)/(5),-(9)/(4))`

Answer

To find the radical center of the given circles represented by the equations:

1. \( S_1: x^2 + y^2 - 2x - 2y + 1 = 0 \)
2. \( S_2: x^2 + y^2 - 4x + 6y - 2 = 0 \)
3. \( S_3: x^2 + y^2 + 3x - 2y + 4 = 0 \)

we will follow these steps:

### Step 1: Write the equations in standard form
We can rewrite the equations of the circles in the standard form by completing the square.

1. For \( S_1 \):
\[ S_1: (x-1)^2 + (y-1)^2 = 1 \]
This represents a circle with center (1, 1) and radius 1.

2. For \( S_2 \):
\[ S_2: (x-2)^2 + (y+3)^2 = 3 \]
This represents a circle with center (2, -3) and radius \(\sqrt{3}\).

3. For \( S_3 \):
\[ S_3: (x+\frac{3}{2})^2 + (y+1)^2 = \frac{25}{4} \]
This represents a circle with center \((-1.5, -1)\) and radius \(\frac{5}{2}\).

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

Calculating \( S_1 - S_2 \):
\[ (x^2 + y^2 - 2x - 2y + 1) - (x^2 + y^2 - 4x + 6y - 2) = 0 \]
This simplifies to:
\[ 2x - 8y + 3 = 0 \quad \text{(Equation 1)} \]

Next, we find the radical axis between \( S_2 \) and \( S_3 \):
\[ S_2 - S_3 = 0 \]
Calculating \( S_2 - S_3 \):
\[ (x^2 + y^2 - 4x + 6y - 2) - (x^2 + y^2 + 3x - 2y + 4) = 0 \]
This simplifies to:
\[ -7x + 8y - 6 = 0 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have two equations:
1. \( 2x - 8y + 3 = 0 \)
2. \( -7x + 8y - 6 = 0 \)

We can solve these equations simultaneously.

From Equation 1:
\[ 2x - 8y = -3 \implies x = 4y - \frac{3}{2} \]

Substituting \( x \) in Equation 2:
\[ -7(4y - \frac{3}{2}) + 8y - 6 = 0 \]
Expanding and simplifying:
\[ -28y + \frac{21}{2} + 8y - 6 = 0 \]
Combining like terms:
\[ -20y + \frac{21}{2} - 6 = 0 \]
\[ -20y + \frac{21 - 12}{2} = 0 \implies -20y + \frac{9}{2} = 0 \]
\[ 20y = \frac{9}{2} \implies y = \frac{9}{40} \]

Now substituting \( y \) back to find \( x \):
\[ x = 4\left(\frac{9}{40}\right) - \frac{3}{2} = \frac{36}{40} - \frac{60}{40} = -\frac{24}{40} = -\frac{3}{5} \]

### Step 4: Conclusion
Thus, the radical center is:
\[ \left(-\frac{3}{5}, \frac{9}{40}\right) \]