The circle $x^{2} + y^{2} + 2 a x + c = 0$ and $x^{2} + y^{2} + 2 b y + c = 0$ touches if

\frac{1}{c^{2}} + \frac{1}{b^{2}} = \frac{1}{a}

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\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c}

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\frac{1}{c^{2}} + \frac{1}{a^{2}} = \frac{1}{b}

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\frac{1}{a^{2}} - \frac{1}{b^{2}} = \frac{1}{c}

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Solution

The correct option is D $\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c}$
Two circles touch each other $c_{1} c_{2} = r_{1} \pm r_{2}$
$∴ \sqrt{a^{2} + b^{2}} = \sqrt{a^{2} - c} \pm \sqrt{b^{2} - c}$
$\Rightarrow a^{2} + b^{2} = a^{2} - c + b^{2} - c \pm 2 \sqrt{(a^{2} - c) (b^{2} - c)}$
$\Rightarrow c^{2} = (a^{2} - c) (b^{2} - c)$
$\Rightarrow c^{2} = a^{2} b^{2} - c (a^{2} + b^{2}) + c^{2}$
$\Rightarrow c (a^{2} + b^{2}) = a^{2} b^{2}$
$\Rightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c}$
Ans: B

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