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.
Find a circle equation on a plane, given 3 points
3dcirclesplane-geometry
Related Solutions
As a first step, I'd move $P_1$ to the origin, so that the points become $P_1=(0,0,0)$, $P_2=(x_2-x_1, y_2-y_1, z_2-z_1)$, $P_3=(x_3-x_1, y_3-y_1, z_3-z_1)$. From there it's just a question of applying a rotation matrix.
A rotation matrix $\mathbf{R}$ which preserves all distances and shapes etc is given by
$$ \mathbf{R}=\left( \begin{array} [c]{ccc}% e_{1}^{x} & e_{2}^{x} & e_{3}^{x}\\ e_{1}^{y} & e_{2}^{y} & e_{3}^{y}\\ e_{1}^{z} & e_{2}^{z} & e_{3}^{z}% \end{array} \right) $$
where the vectors $\left( e_{1}^{\alpha},e_{2}^{\alpha},e_{3}^{\alpha }\right) $ for $\alpha\in\left\{ x,y,z\right\} $ are orthogonal and of unit length, i.e.
$$ \sum_{i=1}^{3}e_{i}^{\alpha}e_{i}^{\beta}=\left\{ \begin{array} [c]{ccc}% 1 & & \alpha=\beta\\ 0 & & \alpha\neq\beta \end{array} \right. .% $$
Applying $\mathbf{R}$, your rotated points $P_{i}^{\prime}=\left( x_{i}^{\prime},y_{i}^{\prime},z_{i}^{\prime}\right) $ are given by
$$ \left( \begin{array} [c]{c}% x_{i}^{\prime}\\ y_{i}^{\prime}\\ z_{i}^{\prime}% \end{array} \right) =\mathbf{R}\left( \begin{array} [c]{c}% x_{i}\\ y_{i}\\ z_{i}% \end{array} \right) $$
I suggest making the top row of $\mathbf{R}$ the unit vector in the direction from $P_{1}$ to $P_{2}$, i.e.
$$ \left( \begin{array} [c]{c}% e_{1}^{x}\\ e_{2}^{x}\\ e_{3}^{x}% \end{array} \right) =\frac{1}{\sqrt{\left( x_{1}-x_{2}\right) ^{2}+\left( y_{1}% -y_{2}\right) ^{2}+\left( z_{1}-z_{2}\right) ^{2}}}\left( \begin{array} [c]{c}% x_{1}-x_{2}\\ y_{1}-y_{2}\\ z_{1}-z_{2}% \end{array} \right) $$
which would mean that the line from $P_{1}$ to $P_{2}$ gets mapped into the $x$-axis.
Working out what you could use for the second and third rows of $\mathbf{R}$ is now just a question of solving some simple linear equations, to make sure that the line from $P_{1}$ to $P_{3}$ has no z component.
To solve for $\left( e_{1}^{z},e_{2}^{z},e_{3}^{z}\right)$, the vector $e_{i}^{z}$ must have a zero dot-product with both $e_{i}^{x}$ and $\left( x_{3}-x_{1},y_{3}-y_{1},z_{3}-z_{1}\right) $, so the equations are
$$ \begin{align*} e_{1}^{x}e_{1}^{z}+e_{2}^{x}e_{2}^{z}+e_{3}^{x}e_{3}^{z} & =0\\ \left( x_{3}-x_{1}\right) e_{1}^{z}+\left( y_{3}-y_{1}\right) e_{2}% ^{z}+\left( z_{3}-z_{1}\right) e_{3}^{z} & =0 \end{align*} $$
Hence
$$ \left( \begin{array} [c]{c}% e_{1}^{z}\\ e_{2}^{z}\\ e_{3}^{z}% \end{array} \right) =\lambda_{z}\left( \begin{array} [c]{c}% \left( z_{3}-z_{1}\right) e_{2}^{x}-\left( y_{3}-y_{1}\right) e_{3}^{x}\\ \left( x_{3}-x_{1}\right) e_{3}^{x}-\left( z_{3}-z_{1}\right) e_{1}^{x}\\ \left( y_{3}-y_{1}\right) e_{1}^{x}-\left( x_{3}-x_{1}\right) e_{2}^{x}% \end{array} \right) $$
for some $\lambda_z$, which should be determined so that $e_{i}^{z}$ is a vector of unit length. Finally $e_{i}^{y}$ can be determined as the vector which is perpendicular to both $e_{i}^{x}$ and $e_{i}^{z}$, i.e.
$$ \left( \begin{array} [c]{c}% e_{1}^{y}\\ e_{2}^{y}\\ e_{3}^{y}% \end{array} \right) =\lambda_{y}\left( \begin{array} [c]{c}% e_{3}^{z}e_{2}^{x}-e_{2}^{z}e_{3}^{x}\\ e_{1}^{z}e_{3}^{x}-e_{3}^{z}e_{1}^{x}\\ e_{2}^{z}e_{1}^{x}-e_{1}^{z}e_{2}^{x}% \end{array} \right) $$
where again $\lambda_{y}$ is determined so that $e_{i}^{y}$ is a vector of unit length.
The answer is yes to both questions. Start with a regular (all faces equilateral triangles) tetrahedron $P_1'P_2'P_3'P_4'$, where the answer is clearly yes to both questions. Then apply an affine transformation $A$ (a combination of stretching, shearing, rotating, and translation) to bring $P_1'P_2'P_3'P_4'$ to $P_1P_2P_3P_4$. For notation, let $A=T\circ L$, where $L$ is linear and $T$ is a translation by some vector $u$.
The transformation will have modified the volume of each of the sub-tetrahedra in the same way: multiplying it by $\det L$. So the sub-tetrahedra of $P_1P_2P_3P_4$ all have the same volume.
The relation that $\frac{P_1'+P_2'+P_3'+P_4'}{4}=O'$ will continue to hold once the transformation is applied to each of these points: $$P_i=AP_i'=LP_i'+u$$
So $$\begin{align} \frac{P_1+P_2+P_3+P_4}{4}&=\frac{LP_1'+LP_2'+LP_3'+LP_4'+4u}{4}\\ &=L\left(\frac{P_1'+P_2'+P_3'+P_4'}{4}\right)+u\\ &=LO'+u\\ &=AO'\\ \frac{P_1+P_2+P_3+P_4}{4}&=O \end{align}$$
Best Answer
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.