Question

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

• A. $x+2y+13=0,2x-y+6=0$
• B. $2x+y+13=0,x-2y=6$
• C. $3x+2y+23=02x-3y+4=0$
• D. $x-7y=23,6x+13y=4$

Answer 1

Answer: (A)

Any line through the point $(-5,-4)$ is
$y+4=m(x+5)$
$mx-y+(5m-4)=0\dots (i)$
Now, radius of circle
$=\sqrt{{\left(2\right)}^2+{\left(3\right)}^2-8}=\sqrt{4+9-8}=\sqrt{5}$
If it is a tangent , then perpendicular from centre $(-2,-3)$ is equal to the above radius.
$\therefore \frac{m\left(-2\right)-\left(-3\right)+\left(5m-4\right)}{\sqrt{m^2+1}}=\sqrt{5}$
$\Rightarrow -2m+3+5m-4=\sqrt{5}\sqrt{1+m^2}$
$\Rightarrow 3m-1=\sqrt{5}\sqrt{1+m^2}$
$\Rightarrow {\left(3m-1\right)}^2=5\left(1+m^2\right)$
$\Rightarrow 9m^2+1-6m=5+5m^2$
$\Rightarrow 4m^2-6m-4=0$
$\Rightarrow 4m^2-8m+2m-4=0$
$\Rightarrow 4m\left(m-2\right)+2\left(m-2\right)=0$
$\Rightarrow \left(m-2\right)\left(4m+2\right)=0$
$\Rightarrow m=2,-\frac{1}{2}$
Putting the value of m = 2 in Eq. (i) , we get
$2x-y+5x2-4=0$
$\Rightarrow 2x-y+6=0$
Again, putting the value of $m=-\frac{1}{2}$ in Eq. (i) , we get
$-\frac{1}{2}x-y+5\left(-\frac{1}{2}\right)-4=0$
$\Rightarrow x-2y-5-8=0$
$\Rightarrow x+2y+13=0$