What is the equation of a general circle in 3-D space?

Question

I know that $(x-x_0)^2+(y-y_0)^2-r^2=0$ is a general planar circle and $(x-x_0)^2+(y-y_0)^2+(z-z_0)^2-r^2=0$ is a general sphere.

I want to know the general expression of a circle in space. Can anyone please give me some advice here?

Answer 1

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Vec}{\mathbf{v}}$A circle in $\Reals^{3}$ doesn't exactly have "an equation" in the sense that a circle in the plane or a sphere in space has an equation. Intuitively, "an equation" cuts down the dimension by one, but to get a circle in space you have to lower the dimension by two. (A small piece of a circle looks like a line, so a circle is "one dimensional" for present purposes.)

Here are a few alternative descriptions that you may find helpful or interesting.

• (Parametric) If $\Vec_{1}$ and $\Vec_{2}$ are orthogonal unit vectors in $\Reals^{3}$, $p$ is an arbitrary point, and $r > 0$ is real, then the set of points of the form $$ p + r(\cos t) \Vec_{1} + r(\sin t) \Vec_{2},\quad \text{$t$ real,} $$ is the circle with center $p$ and radius $r$ lying in the plane parallel to $\Vec_{1}$ and $\Vec_{2}$. (The same technique describes an arbitrary circle in $\Reals^{n}$ for $n \geq 2$.)

• (Intersection of a sphere and a plane) Suppose $p = (x_{0}, y_{0}, z_{0})$ is an arbitrary point, $r > 0$ is real, and $n = (a, b, c)$ is a non-zero vector, then the common solution set of the equations $$ (x - x_{0})^{2} + (y - y_{0})^{2} + (z - z_{0})^{2} = r^{2},\quad a(x - x_{0}) + b(y - y_{0}) + c(z - z_{0}) = 0 $$ is the circle with center $p$ and radius $r$ lying in the plane orthogonal to $n$.

• (Degenerate equation) You can write down a single polynomial in three variables whose zero set is a circle, but the resulting equation is "degenerate" in the sense that every point of the circle is a multiple root of the polynomial. For example, $$ \bigl((x^{2} + y^{2}) - 1\bigr)^{2} + z^{2} = 0 $$ "cuts out" the unit circle in the $(x, y)$-plane of $\Reals^{3}$; the only way the left-hand side can vanish is if each summand vanishes, so that $x^{2} + y^{2} = 1$ and $z = 0$. To do this in general, you can use the equations from the preceding bullet point: $$ \bigl((x - x_{0})^{2} + (y - y_{0})^{2} + (z - z_{0})^{2} - r^{2}\bigr)^{2} + \bigl(a(x - x_{0}) + b(y - y_{0}) + c(z - z_{0})\bigr)^{2} = 0. $$

Answer 2

Hint: A circle is the intersection of a sphere with a plane.

Answer 3

To specify a circle in 3D, you need to know its center, its radius, and also how it's "tilted", which means which plane it lives in. So you will need two equations, one defining the relevant sphere (which specifies the center and radius) and one defining the relevant plane (which specifies the tilt).

Answer 4

The equation of a circle in $3$-space is given by $$ |x|^2 + 2\xi \cdot x + \xi^2 = 0 $$ where $\xi = (\xi_1, \xi_2, \xi_3) \in {\mathbb C}^3$, $x\in {\mathbb R}^3$, $\xi \cdot x = \xi_1x_1+\xi_2x_2+\xi_3x_3$ and $\xi^2 = \xi_1{}^2 + \xi_2{}^2+\xi_3{}^2$.

To see this, write $\xi = - a+ i n$ where $a\in {\mathbb R}^3$ is the centre of the circle and $n\in {\mathbb R}^3$ is the normal to the plane of the circle with $|n| = r$ where $r$ is the radius of the circle.

This representation goes back to Laguerre : E. N. Laguerre, Sur l’emploi des imaginaires dans la g´eom´etrie de l’espace, Nouvelles Annales de Math., Series 2, vol. xi, (1872).

See also P. Baird, Conformal foliations by circles and complex isoparametric functions on Euclidean 3-space. Math. Proc. Cambridge Philos. Soc., 123, (1998) 273–300.

Answer 5

Referring to @Hwang's 2nd Bullet point, above, if you know two of the 3 coordinates of a point on the circle, it is trivial to calculate the 3rd. Solving for the second coordinate in terms of the 1st is just some tedious algebra. Here I show how to solve for $x$ in terms of $y$, with some variable substitutions to keep the equations simple.(It adds mental complexity to reduce the equations' apparent complexity). Given two equations and 3 unknowns, its straightforward to produce an equation of one unknown in terms of a 2nd (substituting to remove the 3rd). This produces an equation of the form $$A(X')^{2}+ BX' + C = 0$$ if we assume the second variable, call it $Y'$, is already chosen. This can be plugged into the quadratic formula to solve for $X'$. In this formula, $$A = 1-(a/c)^{2}, \ C = (Y')^{2}(1-(b/c)^{2})-r^{2}$$ and $$B = \frac{-2abY'}{c^{2}}$$ where $a$, $b$, and $c$ are the $x$, $y$,$ z$ components of the normal vector (or equivalently, the $a$, $b$ ,$c$ of Hwang's 2nd equation in his 2nd bullet point), $r$ is the radius, and $$X' = x-P_{x}, \quad Y'=y-P_{x}.$$

There is an added constraint that when choosing $Y'$, $(Y')^{2} ≤ r^{2}$ which means $$(y-P_{x})^{2} ≤ r^{2}$$ or expressing it another way $$ -r ≤ Y' ≤ +r.$$ (and of course we have the constraints that neither c nor A can be zero) so the equation in its abbreviated form is

$$ X' = \frac{-B ± \sqrt{B²-4AC}}{2A}$$

which will of course generally produce two points (except when $\left(Y'\right)^{2} = r^{2}$).