Position function of a point moving in a circle with velocity $v_0$ and initial position $r_0$ with reflection. Consider the circle given by $x^2+y^2=r^2$ on the x-y axis. Suppose I have a point inside the circle with coordinates $r_0=(x_0, y_0)$. The ball is initially moving in the direction of unit vector $v_0=(a_0, b_0)$ with constant speed of $s_0$. 
Once the ball hits the boundary of the circle, suppose at  at point $(x', \sqrt{r^2-x'^2})$, with an angle $\theta $ to the normal of the point, it reflects back into the circle with the same speed but with angle ($-\theta$) to the normal (i.e. reflection).
Now given only $v_0, r_0,$ and $s_0$, I want to determine the position vector $r$ of the ball at any given time. 
To be honest I don't know how to interact with the different starting points, I can find the formula for some special $r_0$ or $v_0$ but not for the general ones. I have thought about Markov chains but I am not sure if I can use it here. Any help?   
Note: The reason I want to know $r$ is because I want to know if at any point, the ball crosses the center of the circle (i.e. (0,0)). 
 A: As far as I see, it is a ever interesting circular billiard problem.
I could provide some hints, not to ruin the fun.
Thanks to circular simmetry,  the trajectory on circular billiard tables is fully determined by the angle $\theta$.

The  ball trajectory is marked in blue, and is symmetric with respect to the dotted red line, normal to the circumference. One could then for example simply flip the circle, and superimpose pre- and -post bounce trajectory.
There are then immediate consequences on the length of each sub-path between reflections, which dramatically simplify the problem (check https://www.math.psu.edu/tabachni/Books/billiardsgeometry.pdf, Chapter 2, for many interesting hindsights). 
Moreover, if $\theta$ is rational, the path is periodic.
If $\theta$ is irrational, then paths are dense (see again above reference for a rigorous definition and proof).
If your problem is to know whether the ball crosses the centre at any point in time, armed with the information above I believe you are already there.
To cross the centre, the trajectory must have a definite angle with respect to the circumference.
