Find the equation of the circle which cuts the circles $\;$ $x^2 + y^2 - 2x + 3y + 4 = 0$ $\;$ and $\;$ $x^2 + y^2 + 3x - 5y + 1 = 0$ $\;$ orthogonally and which passes through the origin.
Given circles: $\;$ $x^2 + y^2 - 2x + 3y + 4 = 0$ $\;\;\; \cdots \; (1a)$
and $\;$ $x^2 + y^2 + 3x - 5y + 1 = 0$ $\;\;\; \cdots \; (1b)$
Comparing equation $(1a)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0$ $\;\;\; \cdots \; (2a)$ $\;$ gives
$g_1 = -1, \; f_1 = \dfrac{3}{2}, \; c_1 = 4$
Comparing equation $(1b)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0$ $\;\;\; \cdots \; (2b)$ $\;$ gives
$g_2 = \dfrac{3}{2}, \; f_2 = \dfrac{-5}{2}, \; c_2 = 1$
Let the equation of the required circle be
$x^2 + y^2 + 2gx + 2fy + c = 0$ $\;\;\; \cdots \; (2a)$
$\because \;$ Equation $(2a)$ passes through the origin $\left(0, 0\right)$, we have $\;\;\;$ $c = 0$
$\therefore \;$ The equation of the required circle becomes
$x^2 + y^2 + 2gx + 2fy = 0$ $\;\;\; \cdots \; (2b)$
$\because \;$ Circles $(1a)$ and $(2b)$ are orthogonal, we have,
$2gg_1 + 2ff_1 = c + c_1$
i.e. $\;$ $2 \times g \times \left(-1\right) + 2 \times f \times \dfrac{3}{2} = 0 + 4$
i.e. $\;$ $-2g + 3f = 4$ $\;\;\; \cdots \; (3a)$
$\because \;$ Circles $(1b)$ and $(2b)$ are orthogonal, we have,
$2 g g_2 + 2 f f_2 = c + c_2$
i.e. $\;$ $2 \times g \times \dfrac{3}{2} + 2 \times f \times \left(\dfrac{-5}{2}\right) = 0 + 1$
i.e. $\;$ $3g - 5f = 1$ $\;\;\; \cdots \; (3b)$
Solving equations $(3a)$ and $(3b)$ simultaneously gives
$g = -23, \; f = -14$
$\therefore \;$ Equation of the required circle is
$x^2 + y^2 - 46x - 28y = 0$