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

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 $$2y + x + 13 = 0$$Maths

Suggest Corrections

0

Similar questions
View More

People also searched for
View More