Find the equation of the circle which cuts $\;$ $x^2 + y^2 + 4x + 7 = 0$, $\;$ $2x^2 + 2y^2 + 3x + 5y + 9 = 0$ $\;$ and $\;$ $x^2 + y^2 + y = 0$ $\;$ orthogonally.
Given circles:
$x^2 + y^2 + 4x + 7 = 0$ $\;\;\; \cdots \; (1)$
$2x^2 + 2y^2 + 3x + 5y + 9 = 0$
i.e. $\;$ $x^2 + y^2 + \dfrac{3}{2} x + \dfrac{5}{2} y + \dfrac{9}{2} = 0$ $\;\;\; \cdots \; (2)$
$x^2 + y^2 +y = 0$ $\;\;\; \cdots \; (3)$
Comparing equation $(1)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ $\;$ gives
$g_1 = 2, \; f_1 = 0, \; c_1 = 7$
Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0$ $\;$ gives
$g_2 = \dfrac{3}{4}, \; f_2 = \dfrac{5}{4}, \; c_2 = \dfrac{9}{2}$
Comparing equation $(3)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_3 x + 2f_3 y + c_3 = 0$ $\;$ gives
$g_3 = 0, \; f_3 = \dfrac{1}{2}, \; c_3 = 0$
The radical axis of circles $(1)$ and $(2)$ is
$2 x \left(g_1 - g_2\right) + 2y \left(f_1 - f_2\right) + c_1 - c_2 = 0$
i.e. $\;$ $2x \left(2 - \dfrac{3}{4}\right) + 2y \left(0 - \dfrac{5}{4}\right) + 7 - \dfrac{9}{2} = 0$
i.e. $\;$ $\dfrac{5}{2} x - \dfrac{5}{2} y + \dfrac{5}{2} = 0$
i.e. $\;$ $x - y + 1 - 0$ $\;\;\; \cdots \; (4)$
The radical axis of circles $(1)$ and $(3)$ is
$2 x \left(g_1 - g_3\right) + 2y \left(f_1 - f_3\right) + c_1 - c_3 = 0$
i.e. $\;$ $2x \times \left(2 - 0\right) + 2y \left(0 - \dfrac{1}{2}\right) + 7 = 0$
i.e. $\;$ $4x - y + 7 = 0$ $\;\;\; \cdots \; (5)$
Solving equations $(4)$ and $(5)$ simultaneously gives
$x = -2, \; y = -1$
$\therefore \;$ The radical center is $\;$ $\left(h, k\right) = \left(-2, -1\right)$
Square of tangent from $\left(h, k\right)$ to equation $(1)$ is
$h^2 + k^2 + 2 g_1 h + 2 f_1 k + c_1$
$= 4 + 1 + 2 \times 2 \times \left(-2\right) + 2 \times 0 \times \left(-1\right) + 7 = 4$
$\therefore \;$ Required equation of circle is
$\left(x + 2\right)^2 + \left(y + 1\right)^2 = 4$
i.e. $\;$ $x^2 + 4x + 4 + y^2 + 2y + 1 = 4$
i.e. $\;$ $x^2 + y^2 + 4x + 2y + 1 = 0$