Prove that the circles $\;$ $x^2 + y^2 + 2ax +c = 0$ $\;$ and $\;$ $x^2 + y^2 - 2by - c = 0$ $\;$ cut orthogonally.
Given circles: $\;$ $x^2 + y^2 + 2ax + c = 0$ $\;\;\; \cdots \; (1a)$
and $\;$ $x^2 + y^2 - 2by - c = 0$ $\;\;\; \cdots \; (1b)$
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
$g_1 = a, \; f_1 = 0, \; c_1 = c$
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
$g_2 = 0, \; f_2 = -b, \; c_2 = -c$
Circles $(2a)$ and $(2b)$ are orthogonal if
$2g_1 g_2 + 2f_1 f_2 = c_1 + c_2$
$\therefore \;$ We have
$2 \times a \times 0 + 2 \times 0 \times \left(-b\right) = c - c$
i.e. $\;$ $0 = 0$ $\;\;$ which is true.
$\implies$ Circles $(1a)$ and $(1b)$ are orthogonal.