# Determine Circle of Intersection of Plane and Sphere

How can the equation of a circle be determined from the equations of a sphere and a plane which intersect to form the circle? At a minimum, how can the radius and center of the circle be determined?

For example, given the plane equation $$x=\sqrt{3}*z$$ and the sphere given by $$x^2+y^2+z^2=4$$

What is the equation of the circle that results from their intersection? I have used Grapher to visualize the sphere and plane, and know that the two shapes do intersect:

However, substituting $$x=\sqrt{3}*z$$ into $$x^2+y^2+z^2=4$$ yields the elliptical cylinder $$4x^2+y^2=4$$while substituting $$z=x/\sqrt{3}$$ into $$x^2+y^2+z^2=4$$ yields $$4x^2/3+y^2=4$$ Once again the equation of an elliptical cylinder, but in an orthogonal plane.

Why does this substitution not successfully determine the equation of the circle of intersection, and how is it possible to solve for the equation, center, and radius of that circle?

#### Solutions Collecting From Web of "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
$$\Vec{c} = \Vec{c}_{0} + \rho\, \frac{\Vec{n}}{\|\Vec{n}\|} = (x_{0}, y_{0}, z_{0}) + \rho\, \frac{(A, B, C)}{\sqrt{A^{2} + B^{2} + C^{2}}}.$$

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.

You were very close.

y2 = 4 * (1 – x2/3)

y = +/- 2 * (1 – x2/3)1/2 , which gives you two curves

z = x/(3)1/2 (you picked the positive one to plot)

So for a real y, x must be between -(3)1/2 and (3)1/2.