# Circle in polar coordinates ?

May 20, 2016

${\left({y}_{0} - r \cos \left(\theta\right)\right)}^{2} + {\left({x}_{0} - r \sin \left(\theta\right)\right)}^{2} = 1$

#### Explanation:

In Cartesian coordinates, the generic circumference equation with center at point ${p}_{0} = \left({x}_{0} , {y}_{0}\right)$ and radius $r$ is ${\left(x - {x}_{0}\right)}^{2} + {\left(y - {y}_{0}\right)}^{2} = {r}_{0}^{2}$

The pass equations are

$\left(\begin{matrix}x = r \cdot \cos \left(\theta\right) \\ y = r \cdot \sin \left(\theta\right)\end{matrix}\right)$

substituting we have

${\left(r \cdot \cos \left(\theta\right) - {x}_{0}\right)}^{2} + {\left(r \cdot \sin \left(\theta\right) - {y}_{0}\right)}^{2} = {r}_{0}^{2}$.

Simplifying

${r}^{2} - 2 r \left({x}_{0} \cos \left(\theta\right) + {y}_{0} \sin \left(\theta\right)\right) = {r}_{0}^{2} - {x}_{0}^{2} - {y}_{0}^{2}$

If now we call

${x}_{0} = {r}_{0} \cos \left({\theta}_{0}\right)$ and ${y}_{0} = {r}_{0} \sin \left({\theta}_{0}\right)$

and substituting, after the identity

$\cos \left(\theta - {\theta}_{0}\right) = \sin \left(\theta\right) \sin \left({\theta}_{0}\right) + \cos \left(\theta\right) \cos \left({\theta}_{0}\right)$

we get

${r}^{2} + 2 r {r}_{0} \cos \left(\theta - {\theta}_{0}\right) - {r}_{0}^{2} + {x}_{0}^{2} + {y}_{0}^{2}$

Example.

Taking ${\left(x - 5\right)}^{2} + {\left(y - 4\right)}^{2} = 1$
we have ${x}_{0} = 5 , {y}_{0} = 4 , {r}_{0} = 1 , {\theta}_{0} = \arctan \left(\frac{4}{5}\right)$
so

${r}^{2} + 2 r \cos \left(\theta - {\theta}_{0}\right) - 1 + {5}^{2} + {4}^{2} = 0$
or the other presentation
${\left(4 - r \cos \left(\theta\right)\right)}^{2} + {\left(5 - r \sin \left(\theta\right)\right)}^{2} = 1$