3dcircles
Suppose I have a sphere centered at origin. $$ x^2+y^2+z^2=5 $$
and a plane $$ \vec{r}.(\hat{i}+\hat{j}+\hat{k})=3\sqrt{3} $$
And this plane cuts the sphere at a circular region. How do I write the equation of this circle in a 3d Plane. I know its center $center=(\sqrt{3},\sqrt{3},\sqrt{3})$ and its radius $r=4$.
Thanks
$\newcommand{\Vec}[1]{\mathbf{#1}}$Generalities: Let $S$ be the sphere in $\mathbf{R}^{3}$ with center $\Vec{c}_{0} = (x_{0}, y_{0}, z_{0})$ and radius $R > 0$, and let $P$ be the plane with equation $Ax + By + Cz = D$, so that $\Vec{n} = (A, B, C)$ is a normal vector of $P$.
If $\Vec{p}_{0}$ is an arbitrary point on $P$, the signed distance from the center of the sphere $\Vec{c}_{0}$ to the plane $P$ is $$ \rho = \frac{(\Vec{c}_{0} - \Vec{p}_{0}) \cdot \Vec{n}}{\|\Vec{n}\|} = \frac{Ax_{0} + By_{0} + Cz_{0} - D}{\sqrt{A^{2} + B^{2} + C^{2}}}. $$
The intersection $S \cap P$ is a circle if and only if $-R < \rho < R$, and in that case, the circle has radius $r = \sqrt{R^{2} - \rho^{2}}$ and center $$ \Vec{c} = \Vec{c}_{0} + \rho\, \frac{\Vec{n}}{\|\Vec{n}\|} = (x_{0}, y_{0}, z_{0}) + \rho\, \frac{(A, B, C)}{\sqrt{A^{2} + B^{2} + C^{2}}}. $$
Now consider the specific example $$ S = \{(x, y, z) : x^{2} + y^{2} + z^{2} = 4\},\qquad P = \{(x, y, z) : x - z\sqrt{3} = 0\}. $$ The center of $S$ is the origin, which lies on $P$, so the intersection is a circle of radius $2$, the same radius as $S$.
When you substitute $x = z\sqrt{3}$ or $z = x/\sqrt{3}$ into the equation of $S$, you obtain the equation of a cylinder with elliptical cross section (as noted in the OP). However, you must also retain the equation of $P$ in your system. That is, each of the following pairs of equations defines the same circle in space: \begin{align*} x - z\sqrt{3} &= 0, & x - z\sqrt{3} &= 0, & x - z\sqrt{3} &= 0, \\ x^{2} + y^{2} + z^{2} &= 4; & \tfrac{4}{3} x^{2} + y^{2} &= 4; & y^{2} + 4z^{2} &= 4. \end{align*} These may not "look like" circles at first glance, but that's because the circle is not parallel to a coordinate plane; instead, it casts elliptical "shadows" in the $(x, y)$- and $(y, z)$-planes.
Note that a circle in space doesn't have a single equation in the sense you're asking.
We determine the distance of the points $C(5/-3/16)$ and $(5/3/16)$ , which is $6$ and denote it with $d$ ( d is the distance of $C$ from the plane). Then, the radius of the circle is $r'=\sqrt{r^2-d^2}=\sqrt{14^2-6^2}=\sqrt{160}=12.649$
Best Answer
You can represent the 3d circle in parametric form:
1) form a local coordinate system X'Y'Z' on the plane with origin at the circle's center and Z' axis in the same direction as plane's normal.
2) This 3d circle can be represented as
$(x'(t), y'(t), z'(t)) = (rcos(t), rsin(t), 0)$ where r=radius
3) Perform coordinate transformation between coordinate systems X'Y'Z' and XYZ.
I will leave the details about how to find the local coordinate system and how to perform coordinate transformation to yourself.