Find the circle which cuts the circles $\;$ $x^2 + \left(y - b\right)^2 = a^2$, $\;$ $\left(x - c\right)^2 + y^2 = a^2$ $\;$ and $\;$ $x^2 + y^2 = a^2$ $\;$ orthogonally.
Given circles:
$x^2 + \left(y - b\right)^2 = a^2$
i.e. $\;$ $x^2 + y^2 - 2by + b^2 - a^2 = 0$ $\;\;\; \cdots \; (1)$
$\left(x - c\right)^2 + y^2 = a^2$
i.e. $\;$ $x^2 + y^2 -2cx + c^2 - a^2 = 0$ $\;\;\; \cdots \; (2)$
$x^2 + y^2 = a^2$ $\;\;\; \cdots \; (3)$
Comparing equation $(1)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ $\;$ gives
$g_1 = 0, \; f_1 = -b, \; c_1 = b^2 - a^2$
Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0$ $\;$ gives
$g_2 = -c, \; f_2 = 0, \; c_2 = c^2 - a^2$
Comparing equation $(3)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_3 x + 2f_3 y + c_3 = 0$ $\;$ gives
$g_3 = 0, \; f_3 = 0, \; c_3 = - a^2$
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. $\;$ $2x \left(0 + c\right) + 2y \left(-b - 0\right) + b^2 - a^2 - c^2 + a^2 = 0$
i.e. $\;$ $2cx - 2by + b^2 - c^2 = 0$ $\;\;\; \cdots \; (4)$
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. $\;$ $2x \times 0 + 2y \left(-b - 0\right) + b^2 - a^2 + a^2 = 0$
i.e. $\;$ $-2by + b^2 = 0$ $\;\;\; \cdots \; (5)$
From equation $(5)$, if $\;$ $b \neq 0$, $\;$ then $\;$ $y = \dfrac{b}{2}$
Substituting $\;$ $y = \dfrac{b}{2}$ $\;$ in equation $(4)$ gives,
$2cx - 2 b \times \dfrac{b}{2} + b^2 - c^2 = 0$
i.e. $\;$ $2 cx - c^2 = 0$
$\implies$ $x = \dfrac{c}{2}$ $\;$ if $\;$ $c \neq 0$
$\therefore \;$ The radical center is $\;$ $\left(h, k\right) = \left(\dfrac{c}{2}, \dfrac{b}{2}\right)$
Square of tangent from $\left(\dfrac{c}{2}, \dfrac{b}{2}\right)$ to equation $(1)$ is
$h^2 + k^2 + 2 g_1 h + 2 f_1 k + c_1$
$= \dfrac{c^2}{4} + \dfrac{b^2}{4} + 2 \times 0 \times \dfrac{c}{2} + 2 \times \left(-b\right) \times \dfrac{b}{2} + b^2 - a^2$
$= \dfrac{c^2}{4} + \dfrac{b^2}{4} - a^2$
$= \dfrac{b^2 + c^2 - 4 a^2}{4}$
$\therefore \;$ Required equation of circle is
$\left(x - \dfrac{c}{2}\right)^2 + \left(y - \dfrac{b}{2}\right)^2 = \dfrac{b^2 + c^2 - 4a^2}{4}$
i.e. $\;$ $x^2 - cx + \dfrac{c^2}{4} + y^2 - by + \dfrac{b^2}{4} = \dfrac{b^2}{4} + \dfrac{c^2}{4} - a^2$
i.e. $\;$ $x^2 + y^2 - cx - by + a^2 = 0$