The equations of the two tangents from (-5, -4) to the circle $x^{2} + y^{2} + 4 x + 6 y + 8 = 0$ are

x + 2 y + 13 = 0, 2 x - y + 6 = 0

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2 x + y + 13 = 0, x - 2 y = 6

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3 x + 2 y + 23 = 0, 2 x - 3 y + 4 = 0

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x - 7 y = 23, 6 x + 13 y = 4

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Solution

The correct option is A $x + 2 y + 13 = 0, 2 x - y + 6 = 0$
Equation of a line passing through the point $(- 5, - 4)$ is $y = m x + 5 m - 4$
Perpendicular distance of this line from the centre of the given circle, $(- 2, - 3)$ should be equal to the radius, i.e. $\sqrt{2^{2} + 3^{2} - 8}$
$\Rightarrow ∣ ∣ ∣ \frac{- 3 + 2 m - 5 m + 4}{\sqrt{1 + m^{2}}} ∣ ∣ ∣ = \sqrt{5}$

$\Rightarrow | - 3 m + 1 | = \sqrt{5 + 5 m^{2}}$
Squaring both sides, $9 m^{2} - 6 m + 1 = 5 m^{2} + 5$
$\Rightarrow 4 m^{2} - 6 m - 4 = 0$
$\Rightarrow 2 m^{2} - 3 m - 2 = 0$
$\Rightarrow m = \frac{3 \pm \sqrt{9 + 4 \times 2 \times 2}}{4}$
$\Rightarrow m = \frac{- 1}{2}, 2$
The equations of tangents are therefore $y = 2 x + 6$ and $y = \frac{- x}{2} - \frac{13}{2}$
i.e. $2 x - y + 6 = 0$ and $2 y + x + 13 = 0$

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