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Lengths of common tangents of the circles $x^{2} + y^{2} - 2 x + 6 y + 6 = 0; x^{2} + y^{2} + 4 x - 8 y - 5 = 0$ . Are

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1,3

5,3

7,3

5,7

answer is D.

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Detailed Solution

$C_{1} = (1, - 3), r_{1} = 2$

$C_{2} = (- 2, 4), r_{2} = 5$

$d = C_{1} C_{2} = \sqrt{58}$

Length of direct common tangent = $\sqrt{d^{2} - {(r_{1} - r_{2})}^{2}} = 7$

Length of transverse common tangent $= \sqrt{d^{2} - {(r_{1} + r_{2})}^{2}} = 3$

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