Find the equation of the circle which passes through the points of intersection of $\;$ $x^2 + y^2 + 4x - 6y - 12 = 0$ $\;$ and $\;$ $x^2 + y^2 -5x + 7y - 19 = 0$ $\;$ and has its center on $\;$ $x + y = 0$.
The given circles are
$x^2 + y^2 + 4x - 6y - 12 = 0$ $\;\;\; \cdots \; (1)$
$x^2 + y^2 - 5x + 7y - 19 = 0$ $\;\;\; \cdots \; (2)$
Equation of circle through the intersection of equations $(1)$ and $(2)$ is
$\left(x^2 + y^2 + 4x - 6y - 12\right) + k \left(x^2 + y^2 - 5x + 7y - 19\right) = 0$ $\;$ where $k$ is a constant
i.e. $\;$ $\left(1 + k\right) x^2 + \left(1 + k\right) y^2 + \left(4 - 5k\right) x + \left(7k - 6\right) y - 12 - 19k = 0$
i.e. $\;$ $x^2 + y^2 + \left(\dfrac{4 - 5k}{1 + k}\right) x + \left(\dfrac{7k - 6}{1 + k}\right) y + \left(\dfrac{-12 - 19k}{1 + k}\right) = 0$ $\;\;\; \cdots \; (3)$
Comparing equation $(3)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives
$g = \dfrac{4 - 5k}{2 + 2k}, \;\;\; f = \dfrac{7k - 6}{2 + 2k}, \;\;\; c = \dfrac{-12 - 19k}{1 + k}$
Center of the required circle $= \left(-g, -f\right) = \left(\dfrac{5k - 4}{2 + 2k}, \dfrac{6 - 7k}{2 + 2k}\right)$
Center lies on $\;$ $x + y = 0$ $\;\;\; \cdots \; (4)$
$\therefore \;$ We have,
$\dfrac{5k - 4}{2 + 2k} + \dfrac{6 - 7k}{2 + 2k} = 0$
i.e. $\;$ $5k - 4 + 6 - 7k = 0$
i.e. $\;$ $2k = 2$ $\implies$ $k = 1$
Substituting the value of $k$ in equation $(3)$ gives the equation of required circle as
$x^2 + y^2 + \left(\dfrac{4 - 5}{1 + 1}\right)x + \left(\dfrac{7 - 6}{1 + 1}\right)y + \left(\dfrac{-12 - 19}{1 + 1}\right) = 0$
i.e. $\;$ $2x^2 + 2y^2 -x + y - 31 = 0$