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

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Solution

The correct option is A

We know that the equation of common chord is $S_{1} - S_{2}$ = 0, Where $S_{1}$ and $S_{2}$ are the equations of given circles, therefore

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

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

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

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Similar questions

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

(x - a)^{2} + (y - b)^{2} = c^{2}, (x - b)^{2} + (y - a)^{2} = c^{2}, (a \neq b)

(x - a)^{2} + (y - b)^{2} = c^{2}

(x - b)^{2} + (y - a)^{2} = c^{2}

\neq

{(x - a)}^{2} + {(y - b)}^{2} = c^{2}

{(x - b)}^{2} + {(y - a)}^{2} = c^{2}

{(x - a)}^{2} + {(y - b)}^{2} = c^{2} a n d {(x - b)}^{2} + {(y - a)}^{2} = c^{2}

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