I drew from the information provided by Semiclassical and Physicist137 (thank you for helping!) to draw out a direct solution to finding the curve connecting two points.
Suppose we wanted the cycloid connecting an initial, known point $A$ and a second, arbitrary point $B$. For simplicity, set $A=(0,0)$; a different initial point means a simple translation. Using the parametric equations,
\begin{array}{}
x = a(t-\sin t) \\
y = -a(1-\cos t),
\end{array}
where $a$ is the unknown constant or the radius of the rolling circle, we can see that the slope connecting $A$ and $B$ is,
\begin{array}{}
\frac yx = \frac{\cos t+1}{t-\sin t}.
\end{array}
Notice that the slope is independent of the radius $a$. One unknown; one equation. In theory, we can then solve for $t$, $0 \leq t \leq 2\pi$, and then use $t$ to solve for the unknown $a$. From there, you have your parametric equations that would describe the cycloid you're looking for.
To answer my note about the "inverted cycloid" and the inclusion of a negative sign, I realize that in my derivation of the brachistochrone curve, I had defined the $y$-direction -- for gravity -- to be positive downwards. Adjusting for the conventional Cartesian plane where $y$ downwards is negative, you would then swap the signs for $y$; $x$ is unaffected.
I suppose the next question to ask would be about the number of arches you could have connecting both points.
Best Answer
Drawing a diagram is a good start. I don't have the tools for that at the moment though, so we'll have to hope that I can describe this well enough.
$P$ is no longer a point on the circumference of the circle. By generalising we mean that we'll allow $P$ to be somewhere else: either outside or inside the circle. This is where drawing a diagram is a good idea, so you can see how the geometry changes when we move $P$ around. So, let's put $P$ at a distance $\alpha r$ from the centre of the circle, where $\alpha \geq 0$ (you can experiment by letting $\alpha <0$ but it's hard to ascribe a geometric meaning to it then). What changes when we calculate the $x$ and $y$ co-ordinates of $P$?
Well, the distance that the circle moves doesn't change because that's determine by the arc-length of the circumference of the circle. So the term $r\theta$ is unchanged. However, since $P$ is now a distance $\alpha r$ from the centre of the circle, its contribution changes and we have to consider $\sin (\alpha r)$ and $\cos (\alpha r)$ instead.
The generalised parametric form then has co-ordinates $(r\theta - \alpha r \sin \theta, r\theta - \alpha r \cos \theta)$. When you look at the curve that this generates you'll get loops instead of cusps (and, though I'm repeating myself, you should really draw a diagram to appreciate this!)