Equation of Circle/Polar

Theorem
The equation of a circle with radius $R$ and center $\left\langle{r_0, \varphi}\right\rangle$ can be expressed in polar coordinates as:
 * $r^2 - 2 r r_0 \cos \left({\theta - \varphi}\right) + \left({r_0}\right)^2 = R^2$

where:
 * $r_0$ is the distance of the center from the origin
 * $\varphi$ is the angle from the polar axis in the counterclockwise direction
 * $r$ is a function of $\theta$.

Proof
Let the point $\left\langle{r, \theta}\right\rangle_\text {Polar}$ satisfy the equation:
 * $r^2 - 2 r r_0 \cos \left({\theta - \varphi}\right) + \left({r_0}\right)^2 = R^2$

Let the points $\left\langle{r, \theta}\right\rangle$ and $\left\langle{r_0, \varphi}\right\rangle$ be rewritten in Cartesian coordinates:


 * $\left\langle{r, \theta}\right\rangle_\text {Polar} = \left({r \cos \theta, r \sin \theta}\right)_\text{Cartesian}$


 * $\left\langle{r_0, \varphi}\right\rangle_\text{Polar} = \left({r_0 \cos \varphi, r_0 \sin \varphi}\right)_\text{Cartesian}$

Thus the distance between the point $\left\langle{r, \theta}\right\rangle_\text {Polar}$ and the center is:
 * $\sqrt{\left({r \cos \theta - r_0 \cos \varphi}\right)^2 + \left({r \sin \theta - r_0 \sin \varphi}\right)^2}$

So:

But from the equation, this quantity equals $R$.

Therefore the distance between points satisfying the equation and the center is constant and equal to the radius.