The answer to your title question is yes (stereographic projection). However, given the information you have, no, not really, because one of the pieces of information you're given is the length of the arc joining the two points on the circumference (equivalently, the angle), and these notions are essentially baked into the definition of the trig (or inverse trig) functions.
Nevertheless, I'll describe in more detail the stereographic projection, because I think it's a good way of looking at the circle (parametrization using rational functions), and by considering arclengths, as calculated with respect to this parametrization, one essentially 'discovers' the $\arctan$ function (typically, this is described, in what I feel is a rather ad-hoc way, in first-year calculus courses as Weierstrass' half-tangent angle substitution $t=\tan\frac{\theta}{2}$, but I prefer looking at the geometry first and then arriving at the calculus).
Note that by translating the center of the circle, and by rescaling, we may as well assume that we're dealing with a the unit circle centered at the origin.
We'll consider stereographic projection 'from the west'. This means we fix the point $(-1,0)$, the 'west pole' of $S^1$. For any point $(x,y)$ with $x\neq -1$ (i.e other than the west pole), imagine drawing a line through this point, and the west pole. This ray will intersect the $y$-axis at exactly one point. The $y$-coordinate of this point is what we shall call the stereographic projection of the point $(x,y)$ from the west pole. This is an extremely simple geometric approach for identifying the circle minus a point with the real line (and this easily generalizes to higher dimensions).
If we work this word-description out in terms of formulae, we get the following: let $U=\{(x,y)\in S^1\,:\, x\neq -1\}$, and consider $\sigma:U\to \Bbb{R}$,
\begin{align}
\sigma(x,y)=\frac{y}{1+x},
\end{align}
and the inverse is $\sigma^{-1}:\Bbb{R}\to U$,
\begin{align}
\sigma^{-1}(t)=\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right).
\end{align}
Notice the significance of stereographic projection: it has a very simple geometric meaning, it is invertible, and both $\sigma$ and its inverse are rational functions. As examples, for $t=0$, we get $\sigma^{-1}(0)=(1,0)$, for $t=-1$ we get $\sigma^{-1}(-1)=(0,-1)$, and for $t=+1$, we get $\sigma^{-1}(1)=(0,1)$, and so on.
Now, fix two parameter values $-\infty<t_1<t_2<\infty$. Let us calculate the length of the arc joining the points $\sigma^{-1}(t_1)$ and $\sigma^{-1}(t_2)$:
\begin{align}
L_{t_1,t_2}&=\int_{t_1}^{t_2}\|(\sigma^{-1})'(t)\|\,dt
=\int_{t_1}^{t_2}\frac{2}{1+t^2}\,dt,
\end{align}
where the second equality is a straight-forward differentiation. Of course, we would now recognize this immediately as $2[\arctan(t_2)-\arctan(t_1)]$. But, imagine you were never introduced to trigonometry. This would then be a pretty natural way of developing the subject (simple geometry of lines intersecting circles and other lines, and then using calculus to figure out arclengths).
In your question, you're given the coordinates of point $A$ (meaning you know $t_1$ and $\sigma^{-1}(t_1)$), and you know the arclength between $A$ and $B$ (so you know $L$). In order to figure out the coordinates of the point $B$, using this approach, one would have to invert the integral (i.e we've now arrived at a definition for $\tan$).
