Write the condition so that circles $x^{2} + y^{2} + 2 a x + c = 0$ and $x^{2} + y^{2} + 2 b y + c = 0$ touch externally.

a^{- 2} + b^{- 2} = c^{- 2}

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

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a^{- 2} + b^{- 2} = c^{- 1}

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a^{- 2} - b^{- 2} = c^{- 2}

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Solution

The correct option is C $a^{- 2} + b^{- 2} = c^{- 1}$
$C_{1} (- a, 0)$ and $r_{1} = \sqrt{a^{2} - c}$
$C_{2} (0, - b)$ and $r_{2} = \sqrt{b^{2} - c}$
Since, circles touch externally
$C_{1} C_{2} = r_{1} + r_{2}$
$\Rightarrow \sqrt{a^{2} + b^{2}} = \sqrt{a^{2} - c} + \sqrt{b^{2} - c}$
$\Rightarrow a^{2} + b^{2} = a^{2} + b^{2} - 2 c + 2 \sqrt{a^{2} - c} \sqrt{b^{2} - c}$
$\Rightarrow c^{2} = a^{2} b^{2} + c^{2} - c (a^{2} + b^{2})$
$\Rightarrow a^{- 2} + b^{- 2} = c^{- 1}$
Ans: C

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