Length of the common chord of the circles ${(x - 1)}^{2} + {(y + 1)}^{2} = c^{2} a n d {(x + 1)}^{2} + {(y - 1)}^{2} = c^{2}$ is

\frac{1}{2} \sqrt{c^{2} - 2}

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\sqrt{c^{2} - 2}

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2 \sqrt{c^{2} - 2}

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(c + 2)

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Solution

The correct option is D $2 \sqrt{c^{2} - 2}$
The equation of common chord is given by $S_{1} - S_{2} = 0$

$⟹ (x - 1)^{2} + (x + 1)^{2} + (y + 1)^{2} - (y - 1)^{2} = 0$

$⟹ 4 x = 4 y$

$⟹ x = y$

Points of intersection to any circle is given by

$(x - 1)^{2} + (x + 1)^{2} = c^{2}$

$⟹ 2 x^{2} + 2 = c^{2}$

$⟹ x = \pm \sqrt{\frac{c^{2} - 2}{2}}$

Points of intersection are $A (\sqrt{\frac{c^{2} - 2}{2}}, \sqrt{\frac{c^{2} - 2}{2}}) a n d B (- \sqrt{\frac{c^{2} - 2}{2}}, - \sqrt{\frac{c^{2} - 2}{2}})$

Length of AB = $\sqrt{2 (c^{2} - 2) + 2 (c^{2} - 2)} = 2 \sqrt{c^{2} - 2}$

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\neq

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