# Where does the parameterization for a circle in 3D come from?

I was trying to draw circles in 3D in such a way that if I create a list of unit circles varying $\theta$ by some interval, and parameterizing those for $0≤\phi≤2\pi$, I would get a sphere.

$$x = \cos\phi\cos\theta \\ y = \sin\phi \\ z = \cos\phi\sin\theta$$

I stumbled upon a similar question and I'm wondering why it works. I understand of course why a circle would be parameterized in 2D by $(\cos\phi,\sin\phi)$ but I'm not sure why this parameterization tilts the unit circle - specifically how the z-coordinate makes sense.

• Can you please clarify what you mean with "in such a way that if I create a list of unit circles varying $\theta$ by some interval, and parameterizing those for $0≤\phi≤2\pi$, I would get a sphere."? Are you asking why these equations parametrize a sphere? I see no tilted circles. Are you asking how to make tilted circles? – M. Winter Oct 14 '17 at 8:20
• @M.Winter - yes - I was trying to take many unit circles parameterized in 3D, each lying in some different plane $z=n*y$ as $n$ varies from 0 to infinity (in a way, not sure how to better explain it) and superimposing all these circles together each rotated in 3D space, you get a sphere. imgur.com/a/6InMe – rb612 Oct 14 '17 at 8:23
• You mean you want to know why this answer to the other question is working? Maybe leave a comment below this answer and ask for clarification. – M. Winter Oct 14 '17 at 8:29
• @M.Winter well I’ve seen this parameterization before for a unit circle tilted in 3D, so this is more general. – rb612 Oct 14 '17 at 8:54
• Do you know about spherical coordinates and rotation matrices? – M. Winter Oct 14 '17 at 11:00

$2$-dimensional rotations are enough to understand all of this.

$3$-dimensional circles

You said you know that $(\cos\phi,\sin\phi)^\top$ describes a circle in $2$D and you want to consider many rotated copies of this circle so that it formes a sphere. The first thing we have to do is to go to higher dimensions. So when our circle is placed in the $xy$-plane of out $3$-dimensional space, it is given by

$$(\cos\phi,\sin\phi,0)^\top$$

where we just added a zero in the last component to show that it has no expanse into the $z$-direction.

$3$-dimensional rotation matrices

You said you know $2$-dimensional rotation matrices. Here is how to build a simple $3$-dimensional equivalent:

$$R(\phi)=\begin{pmatrix} \cos \phi&\sin\phi\\ -\sin\phi &\cos\phi \end{pmatrix}\quad\Rightarrow\quad R_z(\phi)=\begin{pmatrix} \cos \phi&\sin\phi&0\\ -\sin\phi &\cos\phi&0\\ 0&0&1 \end{pmatrix}.$$

While in two dimensions there is no axis to rotate on (we rotate around points), in three dimensions we have to specify such an axis. This example is very simple. Our modified matrix just has another row/columns which shows that we completely ignore the point's $z$-coordinate. We rotate as if it where a $2$D point in the $xy$-plane. After rotation the point has the same $z$-coordinate as before the rotation (because of the $1$ in the $zz$-entry of the matrix).

So the matrix $R_z(\phi)$ rotates around the $z$-axis by an amount $\phi$. Here are the matrices which rotate around $y$- and $x$-axis respecitively (I think you can see why):

$$R_y(\phi)=\begin{pmatrix} \cos \phi&0&\sin\phi\\ 0&1&0\\ -\sin\phi &0&\cos\phi \end{pmatrix},\qquad R_x(\phi)=\begin{pmatrix} 1&0&0\\ 0&\cos \phi&\sin\phi\\ 0&-\sin\phi &\cos\phi \end{pmatrix}.$$

The rotated circle

Now we have our $3$D circle in the $xy$-plane around the $z$-axis. To make it rotate in such a way that it forms a sphere, we must rotate it around an axis other than $z$, e.g. the $x$-axis. So we take out circle-description $(\cos\phi,\sin\phi,0)^\top$ and we apply a rotation matrix $R_x(\theta)$:

$$\begin{pmatrix} 1&0&0\\ 0&\cos \theta&\sin\theta\\ 0&-\sin\theta&\cos\theta \end{pmatrix}\begin{pmatrix}\cos\phi\\\sin\phi\\0\end{pmatrix} =\begin{pmatrix}\cos\phi\\\sin\phi\cos\theta\\-\sin\phi\sin\theta\end{pmatrix}.$$

Well, this last vector might not be exactly what was given to you by the other answer, but this may be the case because they chose a circle in an other plane than the $xy$-plane or rotated around an other axis than $z$. You can play a bit with this idea to find the correct configuration. At least, this result shows structural similarities to the one given to you in the other answer.