The equation of the circle passing through the intersection of the circles
`x^(2)+y^(2)-6x+2y+4=0`
`x^(2)+y^(2)+2x-4y-6=0`
and having its centre on the line y=x is

Options

A. `3(x^(2)+y^(2))-5x-5y +2=0`
B. `7(x^(2)+y^(2)) -10x -10y-12=0`
C. `x^(2)+y^(2)-2x-2y+1=0`
D. `x^(2)+y^(2)-6x-6y+12=0`

Answer

To find the equation of the circle passing through the intersection of the given circles and having its center on the line \(y = x\), we will follow these steps:

### Step 1: Write the equations of the given circles
The equations of the circles are:
1. \(S_1: x^2 + y^2 - 6x + 2y + 4 = 0\)
2. \(S_2: x^2 + y^2 + 2x - 4y - 6 = 0\)

### Step 2: Form the equation of the circle passing through the intersection
The equation of the circle passing through the intersection of the two circles can be expressed as:
\[ S = S_1 + \lambda S_2 = 0 \]
where \(\lambda\) is a constant.

### Step 3: Substitute the equations into the combined equation
Substituting \(S_1\) and \(S_2\) into the equation:
\[ (x^2 + y^2 - 6x + 2y + 4) + \lambda (x^2 + y^2 + 2x - 4y - 6) = 0 \]
This simplifies to:
\[ (1 + \lambda)x^2 + (1 + \lambda)y^2 + (-6 + 2\lambda)x + (2 - 4\lambda)y + (4 - 6\lambda) = 0 \]

### Step 4: Compare with the general equation of a circle
The general equation of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From our equation, we can identify:
- \(g = \frac{-6 + 2\lambda}{2(1 + \lambda)}\)
- \(f = \frac{2 - 4\lambda}{2(1 + \lambda)}\)
- \(c = \frac{4 - 6\lambda}{1 + \lambda}\)

### Step 5: Center on the line \(y = x\)
For the center of the circle \((-g, -f)\) to lie on the line \(y = x\), we need:
\[ -g = -f \implies g = f \]
Substituting the expressions for \(g\) and \(f\):
\[ \frac{-6 + 2\lambda}{2(1 + \lambda)} = \frac{2 - 4\lambda}{2(1 + \lambda)} \]
This simplifies to:
\[ -6 + 2\lambda = 2 - 4\lambda \]
Rearranging gives:
\[ 6\lambda = 8 \implies \lambda = \frac{4}{3} \]

### Step 6: Substitute \(\lambda\) back into the equation
Substituting \(\lambda = \frac{4}{3}\) back into the combined equation:
\[ (1 + \frac{4}{3})x^2 + (1 + \frac{4}{3})y^2 + (-6 + 2 \cdot \frac{4}{3})x + (2 - 4 \cdot \frac{4}{3})y + (4 - 6 \cdot \frac{4}{3}) = 0 \]
This simplifies to:
\[ \frac{7}{3}x^2 + \frac{7}{3}y^2 + \left(-6 + \frac{8}{3}\right)x + \left(2 - \frac{16}{3}\right)y + \left(4 - 8\right) = 0 \]
\[ \frac{7}{3}x^2 + \frac{7}{3}y^2 - \frac{10}{3}x - \frac{10}{3}y - 4 = 0 \]

### Step 7: Multiply through by 3 to eliminate fractions
Multiplying through by 3 gives:
\[ 7x^2 + 7y^2 - 10x - 10y - 12 = 0 \]

### Final Equation
Thus, the equation of the circle is:
\[ 7x^2 + 7y^2 - 10x - 10y - 12 = 0 \]