Question

The equation of the two tangents from $\left(-5,-4\right)$ 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. $3x+2y+23=0,2x-3y+4=0$
• D. $x-7y=23,6x+13y=4$

Answer 1

The correct option is A

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

Find the equation of two tangents.

Any line through the point $\left(-5,-4\right)$ is $y+4=m\left(x+5\right)$.

$\Rightarrow mx-y+\left(5m-4\right)=0......\left(i\right)$

If it is a tangent, then the perpendicular distance from the centre $\left(-2,-3\right)$ is equal to the radius.

∴Radius

$\begin{array}{rcl}r& =& \sqrt{{\left(x-x_1\right)}^2+{\left(y-y_1\right)}^2}\\ r& =& \sqrt{2^2+3^2-8}\\ & =& \sqrt{5}\end{array}$

The tangent to a circle equation $x^2+y^2=a^2$ for a line $y=mx+c$ is given by the equation $y=mx\pm a\sqrt{[1+m^2]}$

$\begin{array}{rcl}& \Rightarrow & \frac{\left[m-\left(-2\right)-\left(-3\right)+\left(5m-4\right)\right]}{\left[\sqrt{m^2+1}\right]}=\sqrt{5}\\ & \Rightarrow & -2m+3+5m-4=\sqrt{5}\sqrt{m^2+1}\\ & \Rightarrow & {\left(3m-1\right)}^2=\left(5m^2+1\right)\\ & \Rightarrow & 9m^2+1-6m=5m^2+5\\ & \Rightarrow & 4m^2-6m-4=0\\ & \Rightarrow & 2m^2-3m-2=0\\ & \Rightarrow & m=-\frac{1}{2},2\end{array}$

∴ Required equations are $2x-y+6=0$ and $x+2y+13=0$.

Hence, the required equations are $x+2y+13=0,2x-y+6=0$ and therefore option $\left(A\right)$ is correct.