Answer
Since the slope of PQ is equal to the slope of the tangent line at P, the tangent line passes through the top point on the rolling circle.
Work Step by Step
Let the cycloid be generated by the unit circle with parametrization
$c(t)=(t-\sin t,1-\cos t)$.
Using Eq. (8) the slope of the tangent line is
$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\frac{\sin t}{1- \cos t}$.
Let P be the point on the cycloid at $t=t_P$ and Q be the top point of the unit circle. So, the slope of the tangent line at $t_P$ is
$\frac{dy}{dx}|_{t=t_P}=\frac{\sin t_P}{1-\cos t_P}$.
For the unit circle its circumference is $2\pi$, so at $t=t_P$ the unit circle has moved $t_P$ distance to the right of the $y$-axis. The top point of the unit circle is $(t_P,2)$. The slope of PQ is then
$m=\frac{2-\left(1-\cos t_P\right)}{t_P-\left(t_P-\sin t_P\right)}=\frac{1+\cos t_P}{\sin t_P}$.
Multiply $m$ by $\frac{1-\cos t_P}{1-\cos t_P}$ gives
$m=\frac{1-\cos ^2 t_P}{\sin t_P \left(1-\cos t_P\right)}=\frac{\sin ^2 t_P}{\sin t_P\left(1-\cos t_P\right)}=\frac{\sin t_P}{1-\cos t_P}$.
Since the slope of PQ is equal to the slope of the tangent line at P, the tangent line passes through the top point on the rolling circle.
