User:Anghel/Sandbox

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:

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:


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