Analytic expression for the distance between excentric observer and point rotating around center Given two concentric circles with radii $d$ and $d+h$ and a point $A$ moving with constant angular velocity $\omega_0$ around the center $C$.
Point A performs the temporal path
$$A(t) = (d + h) ( \sin(\omega_0 t), \cos(\omega_0 t))$$ in the plane.
I would like to calculate, for the eccentric point $B$ the analytical expression for the distance $d_\text{ex}(t)$ and $\varepsilon_\text{ex}(t)$. However, I'm having difficulties recalling the strategies from back when I was more fit ... ;-)
I don't care about the phase, i.e. $B$ could be anywhere on the inner circle.
I'm also interested in $\frac{d}{dt} d_\text{ex}(t)$, but I'll be able to do this myself.

 A: I assume $\vec{B}(x_b, y_b)$ is fixed. As you've noted, a good parameterization for the location of $A$ is:
$$\vec{A}(t)=(d+h)\cdot (\sin (\omega_0 t), \cos(\omega_0 t))$$
The distance is
$$d_{ex}(t)=||\vec{A}-\vec{B}||=||(d+h)\cdot (\sin (\omega_0 t), \cos(\omega_0 t))-(x_b, y_b)||$$
$$=||((d+h)\cdot \sin(\omega_0t)-x_b, (d+h)\cdot \cos(\omega_0t)-y_b)||$$
$$=\sqrt{((d+h)\cdot \sin(\omega_0t)-x_b)^2+((d+h)\cdot \cos(\omega_0t)-y_b))^2}$$
For a (slightly) more simple differentiation, you might want to open the brackets and get:
$$d_{ex}(t)=\sqrt{(d+h)^2+x_b^2+y_b^2-2(d+h)(y_b\cos(\omega_0 t)+x_b\sin(\omega_0 t))}$$
Since $B$'s distance from the origin is given, we can replace $x_b^2+y_b^2=d^2$, such that
$$d_{ex}(t)=\sqrt{(d+h)^2+d^2-2(d+h)(y_b\cos(\omega_0 t)+x_b\sin(\omega_0 t))}$$
(in polar coordinates, $x_b = d\cos(\theta), y_b = d\sin(\theta)$).
As for the angle between the two vectors, you can find it using the dot product:
$$\require{cancel} \cos (\epsilon_{ex}(t))=\frac{\vec{A}\cdot \vec{B}}{||A||\cdot||B||}=\frac{\cancel{(d+h)}\cdot (\sin (\omega_0 t), \cos(\omega_0 t))\cdot(x_b, y_b)}{\cancel{||(d+h)\cdot (\sin (\omega_0 t), \cos(\omega_0 t))||} \cdot ||(x_b, y_b)||}$$
$$=\frac{x_b\cdot \sin(\omega_0t)+y_b\cdot \cos(\omega_0t)}{d}$$
The cancellation is immediate if we notice that the distance of $A$ from the origin is always $d+h$ (the same argument was used for $B$). The problem here is that $\cos(x)$ is not bijective. Therefore, it doesn't have an inverse that would work for all $x$, and we must look at different cases for $\epsilon_{ex}(t)$.
Notice that for getting rid of either $x_b$ or $y_b$, you can use $x_b^2+y_b^2=d^2$. Another way would be to assume $B$ is at angle $\theta$ from the origin, and then $x_b = d\cos(\theta)$ and $y_b = d\sin(\theta)$. This way also leaves you with one parameter for the location of $B$ (i.e, $\theta$).
