Determine Circle of Intersection of Plane and Sphere

$\newcommand{\Vec}[1]{\mathbf{#1}}$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 $$ \Vec{c} = \Vec{c}_{0} + \rho\, \frac{\Vec{n}}{\|\Vec{n}\|} = (x_{0}, y_{0}, z_{0}) + \rho\, \frac{(A, B, C)}{\sqrt{A^{2} + B^{2} + C^{2}}}. $$

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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.