Find the radical center of the circles $\;$ $x^2 + y^2 + 2x + 2y - 2 = 0$, $\;$ $x^2 + y^2 -4x - 6y + 6 = 0$ $\;$ and $\;$ $x^2 + y^2 + 6x - 4y - 12 = 0$.
Given circles: $\;$ $x^2 + y^2 + 2x + 2y - 2 = 0$ $\;\;\; \cdots \; (1)$
$x^2 + y^2 - 4x - 6y + 6 = 0$ $\;\;\; \cdots \; (2)$
$x^2 + y^2 + 6x - 4y - 12 = 0$ $\;\;\; \cdots \; (3)$
Comparing equation $(1)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives
$g = 1, \; f = 1, \; c = -2$
Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0$ $\;$ gives
$g_1 = -2, \; f_1 = -3, \; c_1 = 6$
Comparing equation $(3)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0$ $\;$ gives
$g_2 = 3, \; f_2 = -2, \; c_2 = - 12$
The radical axis of circles $(1)$ and $(2)$ is
$2 x \left(g - g_1\right) + 2y \left(f - f_1\right) + c - c_1 = 0$
i.e. $\;$ $2x \left(1 + 2\right) + 2y \left(1 + 3\right) - 2 - 6 = 0$
i.e. $\;$ $6x + 8 y - 8 = 0$
i.e. $\;$ $3x + 4y - 4 = 0$ $\;\;\; \cdots \; (4)$
The radical axis of circles $(1)$ and $(3)$ is
$2 x \left(g - g_2\right) + 2y \left(f - f_2\right) + c - c_2 = 0$
i.e. $\;$ $2x \left(1 - 3\right) + 2y \left(1 + 2\right) - 2 + 12 = 0$
i.e. $\;$ $- 4x + 6y + 10 = 0$
i.e. $\;$ $-2x + 3y + 5 = 0$ $\;\;\; \cdots \; (5)$
Solving equations $(4)$ and $(5)$ simultaneously gives,
$x = \dfrac{32}{17}, \; y = \dfrac{-7}{17}$
$\therefore \;$ The radical center is $\;$ $\left(\dfrac{32}{17}, \dfrac{-7}{17}\right)$