Answer
$r^{2}+4r\cos\theta-10r\sin\theta+13=0$
Work Step by Step
Expand the squares,
$x^{2}+4x+4+y^{2}-10y+25=16$
$(x^{2}+y^{2})+4x-10y+13=0$
Apply the conversion formulas: $\left\{\begin{array}{ll}
(x,y)=(r\cos\theta,r\sin\theta) & \\
r^{2}=x^{2}+y^{2}, & \tan\theta=\frac{y}{x}
\end{array}\right.$
$r^{2}+4r\cos\theta-10r\sin\theta+13=0$
