Find a circle equation on a plane, given 3 points Given 3 points $p_1 = (x_1,y_1,z_1), p_2 = (x_2,y_2,z_2) ,p_3 = (x_3,y_3,z_3)$ that lie on the circumference of a circle, is there a way to find the general equation of this circle?
If yes, could someone share this equation here.
 A: See : https://en.wikipedia.org/wiki/Circumscribed_circle
In Cartesian system of coordinates :
$$\vec{P_1}=\left[\begin{matrix}x_1\\y_1\\z_1\end{matrix}\right]\qquad
\vec{P_2}=\left[\begin{matrix}x_2\\y_2\\z_2\end{matrix}\right]\qquad
\vec{P_3}=\left[\begin{matrix}x_3\\y_3\\z_3\end{matrix}\right]\qquad$$
The radius of the circle is :
$$R=\frac12\frac{\parallel\vec{P_1}-\vec{P_2}\parallel\:\parallel\vec{P_2}-\vec{P_3}\parallel\:\parallel\vec{P_3}-\vec{P_1}\parallel}{\parallel(\vec{P_1}-\vec{P_2})\times(\vec{P_2}-\vec{P_3})\parallel}$$
$\times\:$ is the cross product of vectors.
The center of the circle is given by :
$$\vec{P_c}=\alpha\:\vec{P_1}+\beta\:\vec{P_2}+\gamma\:\vec{P_3}$$
$$\alpha=\frac12\frac{\parallel\vec{P_2}-\vec{P_3}\parallel^2(\vec{P_1}-\vec{P_2})\bullet(\vec{P_1}-\vec{P_3})}{\parallel(\vec{P_1}-\vec{P_2})\bullet(\vec{P_2}-\vec{P_3})\parallel^2}$$
$$\beta=\frac12\frac{\parallel\vec{P_1}-\vec{P_3}\parallel^2(\vec{P_2}-\vec{P_1})\bullet(\vec{P_2}-\vec{P_3})}{\parallel(\vec{P_1}-\vec{P_2})\bullet(\vec{P_2}-\vec{P_3})\parallel^2}$$
$$\gamma=\frac12\frac{\parallel\vec{P_1}-\vec{P_2}\parallel^2(\vec{P_3}-\vec{P_1})\bullet(\vec{P_3}-\vec{P_2})}{\parallel(\vec{P_1}-\vec{P_2})\bullet(\vec{P_2}-\vec{P_3})\parallel^2}$$
$\bullet\:$ is the dot product of vectors.
Note for record : In case of a larger number of scattered points, a regression method is given in 
https://fr.scribd.com/doc/31477970/Regressions-et-trajectoires-3D . This is also valid for three points only, but more complicated than the above method, thus less convenient as answer to the OP question.
