# What is the polar form of  x^2 + y^2 = 2x?

Nov 12, 2014

${x}^{2} + {y}^{2} = 2 x$, which looks like:

by plugging in $\left\{\begin{matrix}x = r \cos \theta \\ y = r \sin \theta\end{matrix}\right.$,

$\implies {\left(r \cos \theta\right)}^{2} + {\left(r \sin \theta\right)}^{2} = 2 r \cos \theta$

by multiplying out,

$\implies {r}^{2} {\cos}^{2} \theta + {r}^{2} {\sin}^{2} \theta = 2 r \cos \theta$

by factoring out ${r}^{2}$ from the left-hand side,

$\implies {r}^{2} \left({\cos}^{2} \theta + {\sin}^{2} \theta\right) = 2 r \cos \theta$

by ${\cos}^{2} \theta + {\sin}^{2} \theta = 1$,

$\implies {r}^{2} = 2 r \cos \theta$

by dividing by $r$,

$\implies r = 2 \cos \theta$, which looks like:

As you can see above, ${x}^{2} + {y}^{2} = 2 x$ and $r = 2 \cos \theta$ give us the same graphs.

I hope that this was helpful.