Find the equation of the tangent to the circle $x^2+y^2-6x+4y=12$, which are parallel to the straight line $4x+3y+5=0$.

The tangent of the circle $x^2+y^2-6x+4y=12$ which are parallel to the straight line $4x+3y+5=0$ is
$4x+3y+k=0$ , then r=d
$\sqrt{9+4+12}=\left|\frac{4(3)+3(-2)+k}{\sqrt{16+9}}\right|$

we obtain,
$6+k=\pm 25$,
$k=19$ and $-31.$

therefore, the equations of  tangent of the circle are
$4x+3y+19=0$ and $4x+3y-31=0$