Find the equation of the common chord of the circles:
$(x - a)^{2} + (y - b)^{2} = c^{2}, (x - b)^{2} + (y - a)^{2} = c^{2}, (a \neq b)$

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Solution

Equation of chord chord is given by $S_{1} - S_{2} = 0$

$\Rightarrow (x - a)^{2} + (y - b)^{2} - (x - b)^{2} - (y - a)^{2} = 0$

$\Rightarrow [(x - a)^{2} - (x - b)^{2}] + [(y - b)^{2} - (y - a)^{2}] = 0$

$\Rightarrow (b - a) (2 x - a - b) + (a - b) (2 y - a - b) = 0$

$\Rightarrow 2 x - a - b - (2 y - a - b) = 0$

$\Rightarrow 2 x - 2 y = 0 \Rightarrow x - y = 0$

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The equation of the common chord of the circles $(x - a)^{2} + (y - b)^{2} = c^{2}$ and $(x - b)^{2} + (y - a)^{2} = c^{2}$ is

\neq

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