# Equation of Circle/Parametric

## Theorem

The equation of a circle embedded in the Cartesian plane with radius $R$ and center $\tuple {a, b}$ can be expressed as a parametric equation:

$\begin {cases} x = a + R \cos t \\ y = b + R \sin t \end {cases}$

## Proof

Let the point $\tuple {x, y}$ satisfy the equations:

$x = a + R \cos t$
$y = b + R \sin t$

By the Distance Formula, the distance between $\tuple {x, y}$ and $\tuple {a, b}$ is:

$\sqrt {\paren {\paren {a + R \cos t} - a}^2 + \paren {\paren {b + R \sin t} - b}^2}$

This simplifies to:

$\sqrt {R^2 \cos^2 t + R^2 \sin^2 t} = R \sqrt {\cos^2 t + \sin^2 t}$

Then by Sum of Squares of Sine and Cosine, this distance equals $R$.

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

Let the point $\tuple {x, y}$ lie on the embedded circle with radius $R$ and center $\tuple {a, b}$.

As the distance between $\tuple {x, y}$ and $\tuple {a, b}$ is equal to $R$, it follows that:

 $\ds \paren { x-a }^2 + \paren { y-b }^2$ $=$ $\ds R^2$ Distance Formula $\ds \leadsto \ \$ $\ds \paren { \dfrac { x-a }{ R } }^2 + \paren { \dfrac { y-b }{ R } }^2$ $=$ $\ds 1$ $\ds \leadsto \ \$ $\ds {x_0}^2 + {y_0}^2$ $=$ $\ds 1$ Substitute $x_0 = \dfrac { x-a }{ R }$ and $y_0 = \dfrac { y-b }{ R }$

where $\size {x_0}, \size {y_0} \in \closedint 0 1$, otherwise we would have ${x_0}^2 + {y_0}^2 > 1$.

Cosine of Integer Multiple of Pi shows that $\cos 0 = 1$.

Zeroes of Cosine shows that $\map \cos {\dfrac \pi 2} = 0$.

By Cosine Function is Continuous and the Intermediate Value Theorem, there exists $\theta \in \closedint 0 {\dfrac \pi 2}$ such that $\cos \theta = \size {x_0}$.

Sum of Squares of Sine and Cosine shows that $\cos^2 \theta + \sin^2 \theta = 1$.

It follows that $\sin \theta = \size {y_0}$.

If $x-a$ and $y-b$ are both positive, it follows that:

$\cos \theta = \dfrac { x-a } R , \: \sin \theta = \dfrac { y-b } R$

If $x-a$ is positive and $y-b$ is negative, it follows by Cosine Function is Even and Sine Function is Odd that:

$\map \cos {-\theta} = \dfrac { x-a } R , \: \map \sin {-\theta} = \dfrac { y-b } R$

If $x-a$ is negative and $y-b$ is positive, it follows by Cosine of Supplementary Angle and Sine of Supplementary Angle that:

$\map \cos {\pi-\theta} = \dfrac { x-a } R , \: \map \sin {\pi-\theta} = \dfrac { y-b } R$

Finally, if $x-a$ and $y-b$ are both negative, it follows similarly that:

$\map \cos {-\pi+\theta} = \dfrac { x-a } R , \: \map \sin {-\pi+\theta} = \dfrac { y-b } R$

Combining these four cases, we have found $t \in \R$ such that:

$\cos t = \dfrac { x-a } R , \: \sin t = \dfrac { y-b } R$

We rearrange these equations to get:

$\begin {cases} x = a + R \cos t \\ y = b + R \sin t \end {cases}$

$\blacksquare$