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If $4 l^{2} - 5 m^{2} + 6 l + 1 = 0$ then the $l x + m y + 1 = 0$ touches the circle

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$x^{2} + y^{2} + 6 x - 4 = 0$

$x^{2} + y^{2} - 6 x + 4 = 0$

$x^{2} + y^{2} + 5 x + 4 = 0$

$x^{2} + y^{2} - 2 x + 5 = 0$

answer is B.

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Detailed Solution

$5 m^{2} = 4 l^{2} + 6 l + 1 \Rightarrow 5 (l^{2} + m^{2}) = 9 l^{2} + 6 l + 1$

$\Rightarrow \sqrt{5} = \frac{| 3 l + 1 |}{\sqrt{l^{2} + m^{2}}},$ centre = (3, 0)

$& radius = \sqrt{5}$

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