Equation of Hypocycloid

Theorem
Consider a circle $C_1$ of radius $b$ rolling without slipping around the inside of a circle $C_2$ of (larger) radius $a$ in a cartesian coordinate plane.

Consider a point $P$ on the circumference of $C_1$ where it is tangent to $C_2$ at point $A$ on the $x$-axis.

Consider the hypocycloid traced out by the point $P$.

Let $\left({x, y}\right)$ be the coordinates of $P$ as it travels over the plane.

The point $P = \left({x, y}\right)$ is described by the equations:
 * $x = \left({a - b}\right) \cos \theta + b \cos \left({\left({\dfrac {a - b} b}\right) \theta}\right)$
 * $y = \left({a - b}\right) \sin \theta - b \sin \left({\left({\dfrac {a - b} b}\right) \theta}\right)$

Proof

 * Hypocycloid.png

Let $C_1$ have rolled so that the line $OB$ 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, $P = \left({x, y}\right)$ is defined by:
 * $x = \left({a - b}\right) \cos \theta + b \cos \left({\phi - \theta}\right)$
 * $y = \left({a - b}\right) \sin \theta - b \sin \left({\phi - \theta}\right)$

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:
 * $\phi - \theta = \left({\dfrac {a - b} b}\right) \theta$

whence the result.