# How do you convert -y+2=(3x+4)^2+(y-1)^2 into polar form?

Given that $- y + 2 = {\left(3 x + 4\right)}^{2} + {\left(y - 1\right)}^{2}$
Substituting $x = r \setminus \cos \setminus \theta$ & $y = r \setminus \sin \setminus \theta$
$- r \setminus \sin \setminus \theta + 2 = {\left(3 r \setminus \cos \setminus \theta + 4\right)}^{2} + {\left(r \setminus \sin \setminus \theta - 1\right)}^{2}$
$- r \setminus \sin \setminus \theta + 2 = 9 {r}^{2} \setminus {\cos}^{2} \setminus \theta + 16 + 24 r \setminus \cos \setminus \theta + {r}^{2} \setminus {\sin}^{2} \setminus \theta + 1 - 2 r \setminus \sin \setminus \theta$
$\left(8 \setminus {\cos}^{2} \setminus \theta + 1\right) {r}^{2} + \left(24 \setminus \cos \setminus \theta - \setminus \sin \setminus \theta\right) r + 15 = 0$