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The circles $x^{2} + y^{2} + 2 a x + c = 0 a n d x^{2} + y^{2} + 2 b x + c = 0$ have no common tangent if

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$a b > 0, c < 0$

$a b < 0, c > 0$

$a b > 0, c > 0$

$a b < 0, c < 0$

answer is C.

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Detailed Solution

$C_{1} = (- a, 0) r_{1} = \sqrt{a^{2} - c} C_{2} = (- b, 0) r_{2} = \sqrt{b^{2} - c} If two circles have no common tangents C_{1} C_{2} < |r_{1} - r_{2}| \sqrt{{(a - b)}^{2}} < |\sqrt{a^{2} - c} - \sqrt{b^{2} - c}| ∴ ab > 0, c > 0$

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