Answer:
Option D
Explanation:
We have ,
$x^{2}+y^{2}-6x+4y=12$.
$\Rightarrow$ $(x-3)^{2}+(y+2)^{2}=25$
$\Rightarrow$ $(x-3)^{2}+(y+2)^{2}=5^{2}$
Equation of tangent whose slope m is
$y+2=m(x-3)\pm5\sqrt{m^{2}+1}$ .......(i)
Now, this tangent is parallel to line 4x+3y+5=0
$\therefore$ slope of line is -$\frac{4}{3}$
Put the value of m= -$\frac{4}{3}$ is Eq.(i) , we get
$y+2=-\frac{4}{3} (x-3)\pm 5\sqrt{\left(\frac{-4}{3}\right)^{2}+1}$
$\Rightarrow$ $y+2=\frac{-4}{3} (x-3)\pm 5\left(\frac{5}{3}\right)$
$\Rightarrow$ $3y+6=-4x+12\pm25$
$\Rightarrow$ $4x+3y=6 \pm 25$
$\Rightarrow$ $4x+3y=31$ or $4x+3y=-19$
Hence , equation of tangent is
$4x+3y-31=0 $ or $4x+3y+19=0$
