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Lengths of common tangents of the circles $x^{2} + y^{2} = 6 x, x^{2} + y^{2} + 2 x = 0$ are

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$\sqrt{3}$

$\sqrt{3}, 3 \sqrt{3}$

$2 \sqrt{3}$

$2 \sqrt{3}, 3 \sqrt{3}$

answer is C.

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Detailed Solution

$C_{1} = (3, 0), r_{1} = 3$

$C_{2} = (- 1, 0), r_{2} = 1$

$d = C_{1} C_{2} = 4$

The length of transverse common tangent is $\sqrt{d^{2} - {(r_{1} + r_{2})}^{2}} a n d t h e l e n g t h o f d i r e c t c o m m o n t a n g e n t i s \sqrt{d^{2} - {(r_{1} - r_{2})}^{2}}$

$\sqrt{4^{2} - 4^{2}}, \sqrt{4^{2} - 4}$

$= 0, \sqrt{12}$

$= 0, 2 \sqrt{3}$

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