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The lengh of the transverse common tangent of the circles $x^{2} + y^{2} - 2 x + 4 y + 4 = 0$ and
$x^{2} + y^{2} + 4 x - 2 y + 1 = 0 is$

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$\sqrt{3}$

$\sqrt{17}$

$\sqrt{15}$

answer is D.

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Detailed Solution

$C_{1} = (1 - 2), r_{1} = \sqrt{1 + 4 - 4} = 1; C_{2} = (- 2, 1), r_{2} = \sqrt{4 + 1 - 1} = 2 . d = C_{1} C_{2} = \sqrt{9} + 9 = \sqrt{18}$

The length of transverse common tangent $= \sqrt{d^{2} - {(r_{1} + r_{2})}^{2}}$

$= \sqrt{18 - (1 + 2)^{2}} = \sqrt{18 - 9} = \sqrt{9} = 3$

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