You have to understand that there will be multiple points that satisfy the mathematical constraint here. Think clearly, if you have two points on a sphere (ignore the geodesic form for now), P1 and P2, and you have another point T1, at distance x from P1 and distance y from P2, then you will have another symmetric (mirrored) point T1' which will satisfy the same distance conditions, on the other side, so to speak.
Even worse: Consider the case of a sphere with a diameter D. Your P1 is at the North Pole, and your P2 is at the South pole. Do you see that the all points on the equator will satisfy your condition?
Apply this to your example: Consider P1 at the North Pole. Consider P2 at the South Pole. Consider distance to Px, i.e. D1 = D2 = (2.pie.r)/4. See the problem? All points on the equator satisfy this, not a single unique point. In fact, for this case, even if D1 != D2, then you have smaller concentric lines (concentric to the equator) whose points satisfy these constraints.
Too many Px's in your case, not one. To come to a singularity point on a spherical surface, the description constraints would be more specific.
Lastly, establishing correctness of the context is important. Should your algorithm support all points that meet the criteria? Or should your criteria be altered such that the algorithm evaluates to a singular point, always. Be careful.
Some links to help you:
Wikipedia: http://en.wikipedia.org/wiki/Spherical_coordinates
SO: Plotting a point on the edge of a sphere
Updates, based on your three point example:
Again, there can be multiple points satisfying your criteria. What if P1, P2, P3 lie on the same arc? See the diagram below. Even with three points, there is no guarantee that there will be a single fourth point satisfying the distance criteria. Even with n points, there is no such guarantee.
In mathematical language, for a set of n random points, and a set of distances from these individual points, the set of resulting points that satisfy the distance criteria MAY have more than one elements.
You may be fooled into thinking: Oh, this guy is always assuming points lying on the same arc. Well, you are not making a special algorithm, are you? Your algo will be a generalized solution, won't it?
You need to guarantee that the points are not on the same arc (in a set of n points, I think at least 1 point cannot be on the same arc).
For keeping source points to a bare minimum : You need to establish traingular relations between points, because then, using ONLY two points, the triangle relation will yield exactly one point.
What triangle? Visualize this: You have two points, and a third unknown point. All distance you mention are SPHERICAL, i.e. curved surface distances. Do you see that there are also flat distances between these points? Can you visualize, that there will be a plane passing through these points, slicing the sphere, right? I say this to emphasize that you do not need to worry about surface curvature (hence 3d steradian angles). You can see the underlying 2d triangle, whose unknown vertex will also be the third point on the sphere surface.
I know this maybe very hard for you to visualize, I'll try making a diagram for this. (Can't find any good online tools!).
Lastly, this will be of significant help: Please read carefully.
Taken from Wikipedia: http://en.wikipedia.org/wiki/Great-circle_distance
The great-circle or orthodromic distance is the shortest distance between any two points on the surface of a sphere measured along a path on the surface of the sphere (as opposed to going through the sphere's interior). Because spherical geometry is different from ordinary Euclidean geometry, the equations for distance take on a different form. The distance between two points in Euclidean space is the length of a straight line from one point to the other. On the sphere, however, there are no straight lines. In non-Euclidean geometry, straight lines are replaced with geodesics. Geodesics on the sphere are the great circles (circles on the sphere whose centers are coincident with the center of the sphere).
Some more updates:
Conversion between long, lats to Cartesian coordinates can be done by the Haversine formula. Google it. See here: Converting from longitude\latitude to Cartesian coordinates
and here: http://en.wikipedia.org/wiki/World_Geodetic_System
