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.
To express $\rho$ as a function of $\theta$, you could express $t$ as a function of $\theta$ by inverting
$$\tan\theta=\pm\frac{1-\cos t}{t-\pi-\sin t}$$and plug the result in the expression of $\rho=\sqrt{x^2+y^2}$.
But there is no closed-form expression for $t(\theta)$ because $t$ appears inside and outside of a trigonometric function, making a transcendental equation.
Following @N74's approach, you can write
$$\rho\cos\theta=\arccos(1\pm\rho\sin\theta)-\pi-\sin(\arccos(1\pm\rho\sin\theta)),$$ which is a nasty implicit polar equation.
Best Answer
At $t$ and $t'$ the coordinates repeat: $x(t)=x(t')\land y(t)=y(t')\implies2t-\pi\sin t=2t'-\pi\sin t'\land\cos t=\cos t'$
$\cos t=\cos t'\implies t'=t+k2\pi;\in\mathbb Z-\{0\}\lor (t'=-t+k'2\pi,t\neq n\pi);k,n\in\mathbb Z$
a) $2t-\pi\sin t=2(t+k2\pi)-\pi\sin (t+k2\pi)$
$\sin(t+k2\pi)-\sin t=4k$, no solutions.
b) $2t-\pi\sin t=2(-t+k'2\pi)-\pi\sin (-t+k'2\pi)$
$4t=\pi(4k'+\sin t-\sin(-t+k'2\pi))$
$4t=\pi(4k'+\sin t+\sin t)$
$t/\pi=k'+(1/2)\sin t$
Having as a valid solution (Wolfram Alpha) $t=\pi/2,t'=-\pi/2$ with $k'=0$ Only this value is needed as the function is periodic and th tangents have the same slope at the other intersection points.
Now the slope at $t$ is $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{2-\pi\cos t}{\pi\sin t}$
And at the intersection $\left.\dfrac{dy}{dx}\right|_{t=\pi/2}=2/\pi;\left.\dfrac{dy}{dx}\right|_{t=-\pi/2}=-2/\pi$