My Math Forum centre circle in 3D

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 October 27th, 2013, 09:43 PM #1 Newbie   Joined: Jun 2012 Posts: 13 Thanks: 0 centre circle in 3D Who can calculate the coordinates of the centre of a circle given 3 points P1(x1,y1,z1),P2(x2,y2,z2),P3(x3,y3,z3) Thanks for any response
 October 28th, 2013, 12:40 AM #2 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: centre circle in 3D Let the center of your circle be the point $(a, b, c)$ and its radius $r$. Then you have the three simultaneous equations $(x_i - a)^2 + (y_i - b)^2 + (z_i - c)^2= r^2 \ \cdots (i)$ for $i= 1, 2, 3$. By expanding the squares in the three equations and making pairwise subtractions $(1) - (2) ; \ (1) - (3) ; \ (2) - (3)$, your new set of equations will no longer have $r^2 , a^2 , b^2 , c^2$. Unfortunately, this new system of equations is not linearly independent so you need one more, which is the equation of the plane containing the three points. Find the cross product $(P_2 - P_1) \times (P_3 - P_1)= (\alpha , \beta , \gamma)$ and the plane's equation is $\alpha x + \beta y + \gamma z= k$ where $k$ is uniquely determined by plugging any of $P_1 , P_2 , P_3$ into the plane's equation. Fourth equation is $\alpha a + \beta b + \gamma c= k$. Using this last equation together with two of the pairwise subtractions will suffice in determining $(a, b, c)$ and then you may proceed to determine $r$ if you want.
 October 28th, 2013, 09:29 AM #3 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: centre circle in 3D Just for side info, three points are not sufficient to determine a 2-d sphere because 3 points in $\mathbb{R}^3$ are always co-planar and so may determine a unique circle but infinitely many spheres. You will need a fourth point outside the plane of the other 3 to uniquely determine a sphere. Just fit the fourth point into the algorithm I first gave before the plane equation.

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