Given 3 points how to find the centre and radius of a circle in 3 D?

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To find the center and radius of a circle in 3D given three points, it's crucial to clarify whether these points are on the surface of a sphere or arbitrary points in space. Three points alone are insufficient to define a sphere; at least four points are needed for that purpose. The discussion emphasizes using the perpendicular bisectors of segments formed by pairs of the three points to locate the center. These bisectors will intersect at the circle's center. The approach involves determining the plane through the three points and finding the intersection of the bisectors.
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Given 3 points how to find the centre and radius of a circle in 3 D??

Say the 3 points are (a,b,c),(a1.b1,c1) and (a2,b2,c2)...
 
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Do you mean a sphere? Are these arbitrary points anywhere in the sphere or on the surface?
 
Here is the equation of the points {x_i,y_i} for i = 1,2,3 in terms of the variables x and y:

<br /> \left|<br /> \begin{array}{llll}<br /> x^2+y^2 &amp; x &amp; y &amp; 1 \\<br /> x_1^2+y_1^2 &amp; x_1 &amp; y_1 &amp; 1 \\<br /> x_2^2+y_2^2 &amp; x_2 &amp; y_2 &amp; 1 \\<br /> x_3^2+y_3^2 &amp; x_3 &amp; y_3 &amp; 1<br /> \end{array}<br /> \right|=0<br />
 
Do you mean a sphere? Are these arbitrary points anywhere in the sphere or on the surface?

Three points are not enough to define the sphere. Four will do the trick (unless they will not, but you need four at least). Points inside the sphere do not belong to the sphere. IMHO question is very precise.
 
Borek you are absolutely right. I just wanted to make sure.
 
one might find the plane through the three points, then choosing two pairs of points, find the plane perpendicular bisector of the segments they determine.

these three planes should meet at the center of the circle.
 
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