Center of circle given by three points on sphere For given points $A,B,C$ on a sphere, I need to find the centerpoint $MS$ on the sphere of the sperical circle that has $A,B,C$ on its boundary. By that I mean $d_{greatcircle}(MS,A)=d_{greatcircle}(MS,B)=d_{greatcircle}(MS,C)$. Where $d_{greatcircle}$ denotes the great-circle distance on the sphere.
My idea now would be to calculuate the circle through $A,B,C$ in $R^3$ and after that project the Midpoint $MC$ onto the sphere by normalizing it.
Is this projected point the center of the spherical circle?
Or is there an even easier way to do this?
A second thought I just had which may be even easier: can i just calculate $MS$ as the intersection of the line that goes through the origin and has equal distance to $A,B,C$ and the sphere? (I would get two points, but I could find the circle I'm interested in by some distance argument)

 A: You are essentially asking for the centre of the circle that circumscribes the triangle defined by $A, B$ and $C$. Notice this is much simpler than computing the equation of a circle, as we are working with a much simpler geometrical object: a triangle.
How do you find the circumcentre (i.e., the centre of the circumcircle)?

 Find the intersection of two line bisectors of the segment of the triangle described by $A, B$ and $C$.

How can you project this point onto the sphere?

 Divide by the norm of this point.

Do notice that if this circle has its centre on $(0, 0, 0)$ you can't solve this problem.
Update: more about the geodesic distance from the points to the projected circumcentre

To see that the geodesic distances to the centre of the projected circumcentre from any of the 3 points given, is equivalent to showing that the triangles described by the origin, $O$, a given point (we will refer to it as $X$, where $X$ can either be $A, B$ or $C$) and the projected circumcentre, $G$, are the same triangle.

We will refer to the circumcentre of the triangle described by the points $A, B$ and $C$ as $F$.
Notice that the segments $\overline{OG}$ and $\overline{OX}$ are of length $1$, as they are radii of the sphere. So the problem is reduced to show that all $\overline{GX}$ are of the same length.
Now, let's shift our focus to the triangle $\triangle XFG$, notice that no matter which point is $X$, the angle $\angle XFG$ is of $\frac{\pi}{2}$ radians. The segment $\overline{XF}$ is equal to the radius of the circumcircle, and all these triangles share the segment $\overline{FG}$.
So using the triangle congruent criteria of 2 equal sides and equal angle formed by 2 those sides, all these triangles are congruent. Not only this, but they are congruent and have 2 sides of equal length, so it's not hard to deduce that the third side, $\overline{GX}$, are all of the same length.
Thus, we can deduce that the geodesic distance is the same (proving that the triangles $\triangle XOG$ are congruent means that the angles $\angle XOG$ are all the same, and thus the arcs of circumference have the same length).
