This is my first post on this site, I hope it ends well!
So I have a deceptively simple problem, I need to find the center of a sphere given 3 non-colinear points and the desired radius of the sphere. There should be two possible answers.
Thanks for your reply. I am well aware of the equation for a sphere in 3D space, the issue is finding the right a, b, and c values when given r and the 3 points that are solutions to this equation.
There is one answer if you want the three points to be on a great circle of the sphere. Otherwise, there is a whole line of centers.
If are the three points then the points of the plane that is the perpendicular bisector of are equidistant from .
Same said for with respect to .
The points on the line are equidistant from .
That is a messy problem to solve algebraically.
Hey Plato, thanks for the reply. First off I said it was "deceptively simple" haha, meaning its a simple question to ask but very hard one to answer.
But anyway, I'm really confused by why you think there aren't just two answers. Given just 3 points then yes there would be a line of infinite possible points going through the center of the triangle formed by the points. However, we can specify these points down to just 2, one on either side, because we know the radius of the sphere, right?
Three points will determine a unique circle. Taking a line through the center of that circle, perpendicular to the plane determined by the three points, any point along that line will serve as the center for a sphere containing the given three points.
If, in addition to the three circles, you are given a radius, there exist a center on either side of the plane determined by the three points that will give that radius. That is why there are two such spheres.
circumradius there will be two spheres with that radius containing the three given points.
Note the word any. At the circumcenter the sphere with the circumradius has the three points on a great circle of the sphere.
If you understand vector geometry, on the webpage reference above about halfway down are instructions on finding both the circumradius and the circumcenter.