## Precalculus (6th Edition) Blitzer

The rectangular equation is, ${{\left( x-3 \right)}^{2}}+{{\left( y-2 \right)}^{2}}=13$
Using ${{r}^{2}}={{x}^{2}}+{{y}^{2}}$, we will convert the polar equation to the rectangular equation. Therefore, \begin{align} & r=6\cos \theta +4\sin \theta \\ & {{r}^{2}}=6r\cos \theta +4r\sin \theta \\ & {{x}^{2}}+{{y}^{2}}=6x+4y \end{align} We can further simplify it by completing the square on x and y as, \begin{align} & {{x}^{2}}+{{y}^{2}}=6x+4y \\ & {{x}^{2}}+{{y}^{2}}-6x-4y+9+4=13 \\ & {{\left( x-3 \right)}^{2}}+{{\left( y-2 \right)}^{2}}=13 \end{align}