## Precalculus (6th Edition) Blitzer

$r=4\cos \theta$.
Rectangular equation is ${{\left( x-2 \right)}^{2}}+{{y}^{2}}=4$ …… (I) The relation between polar coordinates and rectangular coordinates can be represented as below: $x=r\cos \theta \ \text{ and }\ y=r\sin \theta$ …… (II) Substitute values of $x\ \text{ and }\ y$ from (II) in (I) to get \begin{align} & {{\left( x-2 \right)}^{2}}+{{y}^{2}}=4 \\ & {{\left( r\cos \theta -2 \right)}^{2}}+{{\left( r\sin \theta \right)}^{2}}=4 \\ & {{r}^{2}}{{\cos }^{2}}\theta +4-4r\cos \theta +{{r}^{2}}{{\sin }^{2}}\theta =4 \\ & {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)-4r\cos \theta =0 \end{align} As, $\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)=1$ From here we get, \begin{align} & {{r}^{2}}-4r\cos \theta =0 \\ & {{r}^{2}}=4r\cos \theta \\ & r=4\cos \theta \end{align} Hence the obtained polar expression is $r=4\cos \theta$.