Given a straight line $y=f(x)=0$. Suppose there is a point $A$ moving along the line ($x, 0$) at a constant speed $v$ from $0$ to the right. Then there is another point $B$ circles around point $A$ at constant speed $\omega$ with constant radius $r$.
Suppose the initial condition at $t=0$, point $A$ is at ($0, 0$), point the $B$ is at ($r, 0$), how can I calculate the expression of the trajectory of point $B$ without using time $t$ as the parameter?
Here is what I tried: I treat the ($x,y$) plane as the complex plane. Then $A$'s trajectory is $Traj(A)=vt+0i$. The relative position of $B$ with respect to $A$ is $Traj_{A}(B)=re^{i\omega t}$. Then the absolute trajectory of $B$ is $Traj(B) = Traj(A) + Traj_A(B)$. Which is $Traj(B)=vt + re^{i\omega t}$. Converting back to ($x, y$) coordinates,
$$ x_B(t)=vt + r\cos(\omega t) \\ y_B(t)=r\sin(\omega t) $$
Now I'm stuck. I don't know how to eliminate $t$ to get a trajectory of $B$ in terms of $y_B=h(x_B)$?
More generally, if $A$'s trajectory $y=f(x)$ is not a straight line, but a polynomial $y=f(x) = \sum_{i=0}^{n}a_n x^n$, $n<4$, how to solve this problem?