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Question

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


A
x+2y+13=0,2xy+6=0
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B
2x+y+13=0,x2y=6
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C
3x+2y+23=0,2x3y+4=0
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D
x7y=23,6x+13y=4
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Solution

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

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

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