# 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?

$\newcommand{\Vec}{\mathbf{#1}}$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.

• Very nice answer, especially the explanation with shadows. Apollonius is smiling in the Mathematician's Paradise... May 10, 2015 at 9:06
• @Georges: Kind words indeed; thank you. :) May 10, 2015 at 12:29
• @AndrewD.Hwang Dear Andrew, Could you please help me with the software which you use for drawing such neat diagrams?
– C.S.
Jan 14, 2017 at 10:56
• @S.C.: I use ePiX. It's easiest to use on *nix (GNU/Linux, Mac OSX, ...), but (I'm told) can be run on Windows in Cygwin. Jan 14, 2017 at 11:52
• @AndrewD.Hwang Hi, can you recommend some books or papers where I can learn more about the method you used? I would appreciate it, thanks. :) Sep 14, 2022 at 1:44

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.

• Solving for y yields the equation of a circular cylinder parallel to the z-axis that passes through the circle formed from the sphere-plane intersection. Is it not possible to explicitly solve for the equation of the circle in terms of x, y, and z? Sep 24, 2014 at 18:02
• There are two y equations above, each gives half of the answer. You supply x, and calculate two y values, and the corresponding z. Notice from y^2 you have two solutions for y, one positive and the other negative. Sep 24, 2014 at 20:29