Answer
$r^{2}-6r\cos\theta+2r\sin\theta+6=0$
Work Step by Step
Expand the squares,
$x^{2}-6x+9+y^{2}+2y+1=4$
$(x^{2}+y^{2})-6x+2y+6=0$
Apply the conversion formulas: $\left\{\begin{array}{ll}
(x,y)=(r\cos\theta,r\sin\theta) & \\
r^{2}=x^{2}+y^{2}, & \tan\theta=\frac{y}{x}
\end{array}\right.$
$r^{2}-6r\cos\theta+2r\sin\theta+6=0$
