Arc Circle Intersection in Euclidean 3D While programming on a GPS based game I came upon the following problem for 3D euclidean space:
Assume we have an arc (given by a sphere center coordinate C and two euclidean coordinates $A$ and $B$ on that sphere, with $AC = AB$ and $A !=B$) and another sphere (given by an arbitrary center coordinate $Q$ and radius $R > 0$). All points given in cartesian coordinates $(x, y, z)$. We will exclude the edge cases where $A, B, C$ are aligned.
Now we search for intersect points $X$ (trivial: $0 <= |X| <= 2$) between the arc and the sphere.
Is there a simple formula with acceptable numerical stability (we have GPS related precision limits anyway) for efficiently computing those two (or fewer) coordinates based on the given 5 parameters $A,B,C,Q,R$?

I don;t know about a formula, but this should be relatively easy: 
First, let's call the arc AB by the name J; I assume that A and B are not antipodal (for then J is not unique) or equal (a trivial case), and that by "the arc from A to B" you mean "the shorted of the two pieces of the unique great circle that passes through A and B. 
In that case, there are at most two points of intersection. 
Look at the function on the first sphere defined by $f(X) = $ distance from $X$ to $Q$. On the arc $J$, this is either monotone or has a single extremum. 
You can then use Golden Section search on $f$ to find the extremum (or to discover that it's monotone). If it's monotone, then bisection will find a point (if any) for which $f(X) = R$. If it's unimodal, Golden Section Search will find a point $D$ on the arc $AB$ such that the function is monotone on $AD$ and on $DB$, and you can apply bisection to each half to find a point for which $f(X) = R$. 
Alternatively, you can parameterize the arc from $A$ to $B$ with a path $\gamma$, subtract $Q$ from everything so that $Q$ is at the origin, and divide everything by $R$ so that the radius of the second sphere is $1$. Then you're looking for values $t$ with $\| \gamma(t) \|^2 = 1$. If you use some sort of constant-speed parameterization, this should turn into a quadratic that you can solve with the quadratic formula, but it's not likely to be pretty. 
