If two circles cut a third circle orthogonally then prove that their radical axis or their common chord will pass through the centre of the third circle.

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Solution

Let $S_{1} = 0$ and $S_{2} = 0$ both cut $S_{3} = 0$ orthogonally.
$∴ 2 g_{1} g_{3} + 2 f_{1} f_{3} = c_{1} + c_{3}$
$2 g_{2} g_{3} + 2 f_{2} f_{3} = c_{2} + c_{3}$
Subtracting, we get
$2 g_{3} (g_{1} - g_{2}) + 2 f_{3} (f_{1} - f_{2}) = c_{1} - c_{2} . . . . (1)$
Again the radical axis of $S_{1} = 0$ and $S_{2} = 0$ is given by $S_{1} - S_{2} = 0$
$∴ 2 x (g_{1} - g_{2}) + 2 y (f_{1} - f_{2}) + c_{1} - c_{2} = 0$ .
It will pass through the centre $(- g_{3}, - f_{3})$ of $S_{3} = 0$
if $- 2 g_{3} (g_{1} - g_{2}) - 2 f_{3} (f_{1} - f_{2}) = - (c_{1} - c_{2})$
or $2 g_{3} (g_{1} - g_{2}) + 2 f_{3} (f_{1} - f_{2}) = c_{1} - c_{2}$ .
Above is true by $(1)$ .

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