Precalculus (6th Edition) Blitzer

The required polar equation is $r=12\cos \theta$.
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 $r=12\cos \theta$.