How could I get the coordinates of the circumscribed circle giving the 3 points coords?

Question

I have the coordinates of 3 points through which, a circle should pass . Having the coordinates of the points in 3D, how could I have the coordinates of the center of circumscribed circle ? Also: if one of the points has some deviations and causes a circumscribed circle couldn't pass through the 3 points, is there a way to determine the required coordinates of the third point in a way that the circle could be constructed to find the deviation in space ?

Answer 1

This formula for the center $O$ of a circle is suitable for both 2d and 3d:

\begin{align} O&= A\cdot \frac{a^2\,(b^2+c^2-a^2)}{((b+c)^2-a^2)(a^2-(b-c)^2)} \\ &+B\cdot \frac{b^2\,(a^2+c^2-b^2)}{((a+c)^2-b^2)(b^2-(a-c)^2)} \\ &+C\cdot \frac{c^2\,(b^2+a^2-c^2)}{((b+a)^2-c^2)(c^2-(b-a)^2)} , \end{align}

where $A,B,C$ are the coordinates (2d or 3d) of the three given points, and $a,b,c$ are the side length of the corresponding $\triangle ABC$.

As for deviations of the location of one point, I can see one specific unstable configuration when e.g. $B\approx C$. For $B=C$ the center is $O=\tfrac12(B+C)$, but any small deviation of the location of, say, the point $C$ along the line $AB$ pushes $O$ to infinity, so you can check such a condition and react somewhat reasonably.

Answer 2

If you have the coordinates three points $A, B,$ and $C,$ you can find the perpendicular bisectors of $AB, AC,$ and $BC$ and then you can find the point of concurrency of the three perpendicular bisectors, which will be the center of the unique circle on which $A, B,$ and $C$ fall.