Tangent to Cycloid passes through Top of Generating Circle

Theorem
Let $C$ be a cycloid generated by the equations:
 * $x = a \paren {\theta - \sin \theta}$
 * $y = a \paren {1 - \cos \theta}$

Then the tangent to $C$ at a point $P$ on $C$ passes through the top of the generating circle of $C$.

Proof
From Tangent to Cycloid, the equation for the tangent to $C$ at a point $P = \tuple {x, y}$ is given by:


 * $(1): \quad y - a \paren {1 - \cos \theta} = \dfrac {\sin \theta} {1 - \cos \theta} \paren {x - a \theta + a \sin \theta}$

From Equation of Cycloid, the point at the top of the generating circle of $C$ has coordinates $\tuple {2 a, a \theta}$.

Substituting $x = 2 a$ in $(1)$:

That is, the tangent to $C$ passes through $\tuple {a \theta, 2 a}$ as was required.