The locus of points from which the lengths of the tangents to the two circles $x^{2} + y^{2} + 4 x + 3 = 0$ and $x^{2} + y^{2} - 6 x + 5 = 0$ are in the ratio $2 : 3$ is a circle with centre

(- 6, 0)

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(6, 0)

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(0, 6)

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(0, - 6)

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Solution

The correct option is C $(- 6, 0)$
Let P(h,k) is an external point.Then length of tangent from P(h,k) to $x^{2} + y^{2} + 4 x = 0 i s L_{1}$
$∴ L_{1} = \sqrt{h^{2} + k^{2} + 4 h}$
And length of tangent from P{h,k) to $x^{2} + y^{2} - 6 x + 5 = 0 i s L_{2}$
$∴ L_{2} = \sqrt{h^{2} + k^{2} - 6 h + 5}$
Given $L_{1} : L_{2} = 2 : 3$
$∴ \frac{L_{1}}{L_{2}} = \frac{2}{3}$
$∴ 3 \sqrt{h^{2} + k^{2} + 4 h} = 2 \sqrt{h^{2} + k h^{2} - 6 h + 5}$
$\Rightarrow 9 h^{2} + 9 k^{2} + 36 h = 4 h^{2} + 4 k^{2} - 24 h + 20$
$\Rightarrow 5 h^{2} + 5 k^{2} + 60 h - 20 = 0$
$\Rightarrow h^{2} + k^{2} + 12 h - 4 = 0$
$\Rightarrow x^{2} + y^{2} + 12 x - 4 = 0$
Centre of locus of P(h,k) is $(- 6, 0)$

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