Find the circle which cuts orthogonally each of the circles $\;$ $x^2 + y^2 + 2x + 4y + 1 = 0$, $\;$ $x^2 + y^2 - 4x + 3 = 0$ $\;$ and $\;$ $x^2 + y^2 + 6y + 5 = 0$.
Let the equation of the required circle be $\;$ $x^2 + y^2 + 2g_1 x + 2 f_1 y + c_1 = 0$ $\;\;\; \cdots \; (1)$
$\begin{aligned}
\text{Equations of given circles are: } & x^2 + y^2 + 2x + 4y + 1 = 0 \;\;\; \cdots \; (2a) \\\\
& x^2 + y^2 - 4x + 3 = 0 \;\;\; \cdots \; (2b) \\\\
& x^2 + y^2 + 6y + 5 = 0 \;\;\; \cdots \; (2c)
\end{aligned}$
Standard equation of circle: $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;\;\; \cdots \; (3)$
Comparing equation $(2a)$ with equation $(3)$ gives
$2g_2 = 2 \implies g_2 = 1$; $\;\;$ $2f_2 = 4 \implies f_2 = 2$; $\;\;$ $c_2 = 1$ $\;\;\; \cdots \; (4a) $
Comparing equation $(2b)$ with equation $(3)$ gives
$2g_3 = -4 \implies g_3 = -2$; $\;\;$ $2f_3 = 0 \implies f_3 = 0$; $\;\;$ $c_3 = 3$ $\;\;\; \cdots \; (4b) $
Comparing equation $(2c)$ with equation $(3)$ gives
$2g_4 = 0 \implies g_4 = 0$; $\;\;$ $2f_4 = 6 \implies f_4 = 3$; $\;\;$ $c_4 = 5$ $\;\;\; \cdots \; (4c) $
Now, circles given by equations $(1)$ and $(2a)$ are orthogonal if $\;$ $2g_1 g_2 + 2 f_1 f_2 = c_1 + c_2$
i.e. $2 \times g_1 \times 1 + 2 \times f_1 \times 2 = c_1 + 1$
i.e. $2g_1 + 4f_1 = c_1 - 1$ $\;\;\; \cdots \; (5a)$
Circles given by equations $(1)$ and $(2b)$ are orthogonal if $\;$ $2g_1 g_3 + 2 f_1 f_3 = c_1 + c_3$
i.e. $2 \times g_1 \times \left(-2\right) + 2 \times f_1 \times 0 = c_1 + 3$
i.e. $-4g_1 = c_1 + 3$ $\implies$ $g_1 = \dfrac{- c_1 - 3}{4}$ $\;\;\; \cdots \; (5b)$
circles given by equations $(1)$ and $(2c)$ are orthogonal if $\;$ $2g_1 g_4 + 2 f_1 f_4 = c_1 + c_4$
i.e. $2 \times g_1 \times 0 + 2 \times f_1 \times 3 = c_1 + 5$
i.e. $6 f_1 = c_1 + 5$ $\implies$ $f_1 = \dfrac{c_1 + 5}{6}$ $\;\;\; \cdots \; (5c)$
$\therefore$ $\;$ In view of equations $(5b)$ and $(5c)$ equation $(5a)$ becomes
$2 \times \left(\dfrac{-c_1 - 3}{4}\right) + 4 \times \left(\dfrac{c_1 + 5}{6}\right) = c_1 + 1$
i.e. $\dfrac{-c_1 - 3}{2} + \dfrac{2 \left(c_1 + 5\right)}{3} = c_1 + 1$
i.e. $-3c_1 - 9 + 4 c_1 + 20 = 6 c_1 + 6$
i.e. $5 c_1 = 5$ $\implies$ $c_1 = 1$ $\;\;\; \cdots \; (6a)$
Substituting the value of $c_1$ in equation $(5b)$ gives
$g_1 = \dfrac{-1 - 3}{4} = -1$ $\;\;\; \cdots \; (6b)$
Substituting the value of $c_1$ in equation $(5c)$ gives
$f_1 = \dfrac{1 + 5}{6} = 1$ $\;\;\; \cdots \; (6c)$
$\therefore$ $\;$ In view of equations $(6a)$, $(6b)$ and $(6c)$, equation $(1)$ becomes
$x^2 + y^2 - 2x + 2y + 1 = 0$ $\;\;\; \cdots \; (7)$
Equation $(7)$ gives the required circle orthogonal to the three given circles.