Tangent to Cycloid passes through Top of Generating Circle

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

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 $P = \left({x, y}\right)$ is given by:


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

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

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

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