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)
Best Answer
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)?
How can you project this point onto the sphere?
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).