Answer
The required polar equation is $r=12\cos \theta $.
Work Step by Step
To find the polar equation, substitute $x=r\cos \theta \ \text{ and }\ y=r\sin \theta $ in the provided equation, where r is the distance of that point from the origin and $\theta $ is the respective angle.
So,
$\begin{align}
& {{\left( x-6 \right)}^{2}}+{{y}^{2}}=36 \\
& {{\left( r\cos \theta -6 \right)}^{2}}+{{\left( r\sin \theta \right)}^{2}}=36 \\
& {{r}^{2}}{{\cos }^{2}}\theta +36-12r\cos \theta +{{r}^{2}}{{\sin }^{2}}\theta =36 \\
& {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)-12r\cos \theta =0
\end{align}$
Since, ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ so,
$\begin{align}
& {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)-12r\cos \theta =0 \\
& {{r}^{2}}-12r\cos \theta =0 \\
& r\left( r-12\cos \theta \right)=0 \\
& r=0,r=12\cos \theta
\end{align}$
Because, $r=0$ is a single point. Therefore, the polar form of the equation ${{\left( x-6 \right)}^{2}}+{{y}^{2}}=36$ is $r=12\cos \theta $.
Hence, the required polar equation is $r=12\cos \theta $.
