Find the equation of the circle passing through the point of intersection of the circles `x^2 + y^2 - 6x + 2y + 4 = 0, x^2 + y^2 + 2x - 4y -6 = 0` and with its centre on the line `y = x.`

Answer

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

### Step 1: Identify the equations of the circles
The given circles are:
1. \(C_1: x^2 + y^2 - 6x + 2y + 4 = 0\)
2. \(C_2: x^2 + y^2 + 2x - 4y - 6 = 0\)

### Step 2: Use the formula for the circle passing through the intersection of two circles
The equation of the circle passing through the intersection of two circles can be expressed as:
\[ S_1 + \lambda S_2 = 0 \]
where \(S_1\) and \(S_2\) are the left-hand sides of the equations of the circles.

### Step 3: Substitute the equations into the formula
Substituting the equations of \(C_1\) and \(C_2\):
\[ (x^2 + y^2 - 6x + 2y + 4) + \lambda (x^2 + y^2 + 2x - 4y - 6) = 0 \]
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: Rearranging the equation
Rearranging gives us:
\[ (1 + \lambda)x^2 + (1 + \lambda)y^2 + (-6 + 2\lambda)x + (2 - 4\lambda)y + (4 - 6\lambda) = 0 \]

### Step 5: Identify the center of the circle
The center of the circle can be found using the coefficients of \(x\) and \(y\):
- The center \((h, k)\) is given by:
\[ h = -\frac{-6 + 2\lambda}{2(1 + \lambda)} = \frac{6 - 2\lambda}{2(1 + \lambda)} = \frac{3 - \lambda}{1 + \lambda} \]
\[ k = -\frac{2 - 4\lambda}{2(1 + \lambda)} = \frac{4\lambda - 2}{2(1 + \lambda)} = \frac{2\lambda - 1}{1 + \lambda} \]

### Step 6: Set the center on the line \(y = x\)
Since the center lies on the line \(y = x\), we set \(h = k\):
\[ \frac{3 - \lambda}{1 + \lambda} = \frac{2\lambda - 1}{1 + \lambda} \]
Cross-multiplying gives:
\[ (3 - \lambda) = (2\lambda - 1) \]
Solving this equation:
\[ 3 - \lambda = 2\lambda - 1 \implies 3 + 1 = 2\lambda + \lambda \implies 4 = 3\lambda \implies \lambda = \frac{4}{3} \]

### Step 7: Substitute \(\lambda\) back into the circle equation
Now substitute \(\lambda = \frac{4}{3}\) back into the circle 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:
\[ \frac{7}{3}x^2 + \frac{7}{3}y^2 + (-\frac{14}{3})x + (-\frac{10}{3})y + (-\frac{8}{3}) = 0 \]
Multiplying through by 3 to eliminate the fractions:
\[ 7x^2 + 7y^2 - 14x - 10y - 8 = 0 \]

### Final Equation
Thus, the equation of the required circle is:
\[ 7x^2 + 7y^2 - 14x - 10y - 8 = 0 \]