Is there a nice closed form for inversion on the boundary of an orthogonal circle?

I have two circles $A$ and $B$, such that $A \perp B$.

Let's say I have some point $a\in A$, I'm interested in where the inversion of $a$ over $B$, which I will call $a'$ is. Now since $A$ and $B$ are orthogonal this point will be on $A$ as well.

Additionally we can very easily give classification for $4$ points. The two intersections of $A$ and $B$ are fixed points, and the two points which are equidistant from these intersections are each others inverses.

For other points I can use a compass and straight edge to manually find their inversion quite easily. But I would like to work algebraically.

I'm wondering if there is some nice closed form for this. We can w.l.o.g. fix the size and location of $A$ as well as one of the intersection points between $A$ and $B$. The nrelationship between $A$ and $B$ can be represented as the angle between the two intersections, and $a$ can be represented as the angle from the fixed intersection point.

So I'm looking for a function which takes these two angles and spits out a third angle representing the angle of $a'$ from $x_0$.

By definition, if the center of inversion is $O$, the inverse $P'$ of a point $P$ is in the ray $\overrightarrow{OP}$. In your particular case, $a'$ is just the second intersection of that ray with $A$, hence if you have the coordinates of the center of $B$, the equation of $A$ and the coordinates of $a$, you can easily find $A'$.
There's also a formula for inversion in analytic geometry. Finally, there's also an inversion formula using complex numbers.

I hope this helps.