# Equation of Epicycloid

## Theorem

Let a circle $C_1$ of radius $b$ roll without slipping around the outside of a circle $C_2$ of radius $a$.

Let $C_2$ be embedded in a cartesian plane with its center $O$ located at the origin.

Let $P$ be a point on the circumference of $C_1$.

Let $C_1$ be initially positioned so that $P$ is its point of tangency to $C_2$, located at point $A = \tuple {a, 0}$ on the $x$-axis.

Let $H$ be the epicycloid traced out by the point $P$.

Let $\tuple {x, y}$ be the coordinates of $P$ as it travels over the plane.

The point $P = \tuple {x, y}$ is described by the equations:

$x = \paren {a + b} \cos \theta - b \map \cos {\paren {\dfrac {a + b} b} \theta}$
$y = \paren {a + b} \sin \theta - b \map \sin {\paren {\dfrac {a + b} b} \theta}$

## Proof

Let $C_1$ have rolled so that the line $OC$ through the radii of $C_1$ and $C_2$ is at angle $\theta$ to the $x$-axis.

Let $C_1$ have turned through an angle $\phi$ to reach that point.

By definition of sine and cosine, $P = \tuple {x, y}$ is defined by:

$x = \paren {a + b} \cos \theta - b \map \cos {\phi + \theta}$
$y = \paren {a + b} \sin \theta - b \map \sin {\phi + \theta}$

The arc of $C_1$ between $P$ and $B$ is the same as the arc of $C_2$ between $A$ and $B$.

Thus by Arc Length of Sector:

$a \theta = b \phi$

Thus:

$\phi + \theta = \paren {\dfrac {a + b} b} \theta$

whence the result.

$\blacksquare$