Prove that the length of the common chord of the circles $\left(x - a\right)^2 + \left(y - b\right)^2 = c^2$ and $\left(x - b\right)^2 + \left(y - a\right)^2 = c^2$ is $\sqrt{4c^2 - 2 \left(a - b\right)^2}$.
Given circles:
$\left(x - a\right)^2 + \left(y - b\right)^2 = c^2$
i.e. $\;$ $x^2 + y^2 -2ax -2by + a^2 + b^2 - c^2 = 0$ $\;\;\; \cdots \; (1)$
$\left(x - b\right)^2 + \left(y - a\right)^2 = c^2$
i.e. $\;$ $x^2 + y^2 -2bx - 2ay + a^2 + b^2 - c^2 = 0$ $\;\;\; \cdots \; (2)$
Comparing equation $(1)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_1 x + 2f_1 y + c_1 = 0$ $\;$ gives
$g_1 = -a, \; f_1 = -b, \; c_1 = a^2 + b^2 - c^2$
Comparing equation $(2)$ with the standard equation of circle $\;$ $x^2 + y^2 + 2g_2 x + 2f_2 y + c_2 = 0$ $\;$ gives
$g_2 = -b, \; f_2 = -a, \; c_2 = a^2 + b^2 - c^2$
Common chord of circles $(1)$ and $(2)$ is
$2x \left(g_1 - g_2\right) + 2y \left(f_1 - f_2\right) + c_1 - c_2 = 0$
i.e. $\;$ $2x \left(-a + b\right) + 2y \left(-b + a\right) + a^2 + b^2 - c^2 - a^2 - b^2 + c^2 = 0$
i.e. $\;$ $2 \left(b - a\right)x + 2 \left(a - b\right) y = 0$
i.e. $\;$ $\left(b - a\right) x = \left(b - a\right) y$
i.e. $\;$ $x = y$ $\;\;\; \cdots \; (3)$ $\;\;$ provided $b - a \neq 0$
Substituting $x = y$ in equation $(1)$ gives
$y^2 + y^2 - 2ay - 2by + a^2 + b^2 - c^2 = 0$
i.e. $\;$ $2y^2 - \left(2a + 2b\right) y + a^2 + b^2 - c^2 = 0$
$\begin{aligned}
\therefore \; y & = \dfrac{\left(2a + 2b\right) \pm \sqrt{\left(2a + 2b\right)^2 - 4 \times 2 \times \left(a^2 + b^2 - c^2\right)}}{2 \times 2} \\\\
& = \dfrac{\left(2a + 2b\right) \pm \sqrt{4a^2 + 4b^2 + 8ab - 8a^2 - 8b^2 + 8c^2}}{4} \\\\
& = \dfrac{\left(2a + 2b\right) \pm \sqrt{8c^2 - \left(4a^2 + 4b^2 - 8ab\right)}}{4} \\\\
& = \dfrac{2 \left(a + b\right) \pm 2 \sqrt{2c^2 - \left(a^2 + b^2 - 2ab\right)}}{4} \\\\
& = \dfrac{a + b \pm \sqrt{2c^2 - \left(a - b\right)^2}}{2}
\end{aligned}$
$\therefore \;$ From equation $(3)$,
when $\;$ $y = \dfrac{a + b + \sqrt{2c^2 - \left(a - b\right)^2}}{2}$, $\;$ $x = \dfrac{a + b + \sqrt{2c^2 - \left(a - b\right)^2}}{2}$ $\;$ and
when $\;$ $y = \dfrac{a + b - \sqrt{2c^2 - \left(a - b\right)^2}}{2}$, $\;$ $x = \dfrac{a + b - \sqrt{2c^2 - \left(a - b\right)^2}}{2}$
$\therefore \;$ Coordinates of the end points of the common chord of the two circles are
$P \left(x_1, y_1\right) = \left(\dfrac{a + b + \sqrt{2c^2 - \left(a - b\right)^2}}{2}, \dfrac{a + b + \sqrt{2c^2 - \left(a - b\right)^2}}{2}\right)$ $\;$ and
$Q \left(x_2, y_2\right) = \left(\dfrac{a + b - \sqrt{2c^2 - \left(a - b\right)^2}}{2}, \dfrac{a + b - \sqrt{2c^2 - \left(a - b\right)^2}}{2}\right)$
$\therefore \;$ Square of length of common chord of the two given circles is
$PQ^2 = \left(\dfrac{a + b}{2} + \dfrac{\sqrt{2c^2 - \left(a - b\right)^2}}{2} - \dfrac{a + b}{2} + \dfrac{\sqrt{2c^2 - \left(a - b\right)^2}}{2}\right)^2$
$\hspace{1.5cm}$ $+ \left(\dfrac{a + b}{2} + \dfrac{\sqrt{2c^2 - \left(a - b\right)^2}}{2} - \dfrac{a + b}{2} + \dfrac{\sqrt{2c^2 - \left(a - b\right)^2}}{2}\right)^2$
i.e. $\;$ $PQ^2 = 2c^2 - \left(a - b\right)^2 + 2c^2 - \left(a - b\right)^2$
i.e. $\;$ $PQ^2 = 4c^2 - 2 \left(a - b\right)^2$
$\therefore \;$ Length of common chord of the two given circles is
$PQ = \sqrt{4c^2 - 2 \left(a - b\right)^2}$
Hence proved.