Equation for a circle about some rotation axis in 3D

Question

If, in 3D, I have some rotational axis defined by a theta and phi, how can I describe/find an equation for a circle around such an axis?

The circle is a small circle of the unit sphere.

Answer 1

First calculate the vector representing the axis. This would be

$$\vec a=(\cos\theta\cos\phi,\;\sin\theta\cos\phi,\;\sin\phi)$$

or something similar, depending on the coordinate system. (I use $\theta$ for longitude and $\phi$ for latitude.)

The plane perpendicular to $\vec a$, at distance $b$ from the origin, is the set of all points $\vec r=(x,y,z)$ that satisfy

$$\vec r\cdot\vec a=b$$

(using the vector dot product). The circle is the intersection of this plane with the unit sphere, which means also

$$\vec r\cdot\vec r=x^2+y^2+z^2=1.$$

Answer 2

@mr_e_man's answer is great, and answers the question as asked, but folks also sometimes want a parametric description of sets like these, i.e., something you can work with to explicitly get all points on the set. Let me address that.

$$ \newcommand{vv}{{\mathbf v}} \newcommand{vw}{{\mathbf w}} \newcommand{va}{{\mathbf a}} \newcommand{vx}{{\mathbf x}} $$

Once again, using mr_e_man's notation, I need to split into two cases. The easy case is when the axis is either the north or the south pole, i.e., $\phi = \pm \pi/2$. In this case, the circle is given by $$ c(t) = (\cos t, \sin t, 0), ~~ 0 \le t < 2\pi. $$ When the axis $\va$ is not one of those poles, I'll create two other useful vectors; we let $$ \vv = (\sin \theta, -\cos \theta, 0) $$ and $$ \vw = (\cos \theta \sin \phi, \sin \theta \sin \phi, -\cos \phi) $$ We can then parameterize the curve as follows: $$ c(t) = \cos (t) \vv + \sin (t) \vw. $$

If we think of the axis $a$ as defining a "new" north-south polar line, and the circle $t \mapsto c(t)$ as defining a new equator, then a small circle at distance $d$ from the equatorial plane is given by

$$ q(t) = \sqrt{1-d^2}\cos (t) \vv + \sqrt{1-d^2}\sin (t) \vw + d \va $$ which only makes sense for $-1 \le d \le 1$, of course, for at $d = 1$, we have a circle of zero radius right at the "new" north pole, $\va$.