The equations of the two tangents from (-5, - | vedclass

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(A) Any line passing through the point $(-5, -4)$ with slope $m$ is given by $y+4 = m(x+5)$,which simplifies to $mx - y + (5m - 4) = 0$. For the circl

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

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(A) Any line passing through the point $(-5, -4)$ with slope $m$ is given by $y+4 = m(x+5)$,which simplifies to $mx - y + (5m - 4) = 0$. For the circle $x^{2}+y^{2}+4x+6y+8=0$,the center is $(-2, -3)$ and the radius $r = \sqrt{(-2)^{2} + (-3)^{2} - 8} = \sqrt{4+9-8} = \sqrt{5}$. Since the line is a tangent,the perpendicular distance from the center $(-2, -3)$ to the line must equal the radius: $\frac{|m(-2) - (-3) + (5m - 4)|}{\sqrt{m^{2} + (-1)^{2}}} = \sqrt{5}$ $\frac{|3m - 1|}{\sqrt{m^{2} + 1}} = \sqrt{5}$ Squaring both sides: $(3m - 1)^{2} = 5(m^{2} + 1)$ $9m^{2} - 6m + 1 = 5m^{2} + 5$ $4m^{2} - 6m - 4 = 0$ $2m^{2} - 3m - 2 = 0$ $(2m + 1)(m - 2) = 0$ So,$m = 2$ or $m = -\frac{1}{2}$. For $m = 2$: $2x - y + (5(2) - 4) = 0 \Rightarrow 2x - y + 6 = 0$. For $m = -\frac{1}{2}$: $-\frac{1}{2}x - y + (5(-\frac{1}{2}) - 4) = 0$ $\Rightarrow -x - 2y - 5 - 8 = 0$ $\Rightarrow x + 2y + 13 = 0$.

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

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