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. $3x + 2y + 23 = 0 \, \, \, \, \, 2x - 3y + 4 = 0$
• D. $x-7y = 23, \, \, \, \, \, \, 6x+13y = 4$

Answer 1

Answer: (A)

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

Answer 2

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 E (i) , we get
$2x - y + 5 x 2 - 4 = 0$
$\Rightarrow\, 2x - y + 6 = 0$
Again, putting the value of $m=-\frac{1}{2}$ in E (i) , we get
$-\frac{1}{2}x-y+5\left(-\frac{1}{2}\right)-4=0$
$\Rightarrow x - 2y - 5 - 8 = 0$
$\Rightarrow x + 2y + 13= 0$