Answer
The rectangular equation is, ${{\left( x-3 \right)}^{2}}+{{\left( y-2 \right)}^{2}}=13$
Work Step by Step
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,
$\begin{align}
& {{x}^{2}}+{{y}^{2}}=6x+4y \\
& {{x}^{2}}+{{y}^{2}}-6x-4y+9+4=13 \\
& {{\left( x-3 \right)}^{2}}+{{\left( y-2 \right)}^{2}}=13
\end{align}$
