A cycloid that goes through the beginning and through a general point

Question

Parametric equations of the general cycloid through the beginning $(0,0)$ are

$$x(t)=\frac{2t-\sin2t}{2d}$$

$$y(t)=\frac{1-\cos 2t}{2d}$$

How can we determine $d$ such that the cycloid goes through the point $(a,b)$?

Attempt: Assume for simplicity that $(a,b)=(1,1)$. We are looking for $t$ such that $x(t)=y(t)=1$. This gives us $d=t-\frac12\sin2t$ and $d=\frac12(1-\cos 2t)$. How should we proceed?

Answer 1

First eliminate $d$:

$$\frac{2t-\sin2t}{1-\cos2t}=\frac ab$$

Solve that for $t$. You will have to do so numerically, since this is a transcendental equation. Once you have a suitable $t$ (there might be several to choose from, or even infinitely many), computing $d$ is easy.

Answer 2

Since you received MvG's answer, I give you a second way which is not as good.

As you wrote, you have two equations for two unknowns $d$ and $t$. Using your definitions and requirements, we have $$a=\frac{2t-\sin2t}{2d}$$ $$b=\frac{1-\cos 2t}{2d}$$ From the second equation, we can extract $$t=\frac{1}{2} \cos ^{-1}(1-2 b d)$$ and replace $t$ in the first equation. So, what remains is a quite unpleasant single equation in $d$.