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

$x - y = 0$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$x + y = 0$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$x + y = a^{2} + b^{2}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$x - y = a^{2} - b^{2}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
$x - y = 0$

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

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

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

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

Suggest Corrections

Similar questions

(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}

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

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

\sqrt{4 c^{2} - 2 (a - b)^{2}}

Join BYJU'S Learning Program