First I am a newbie in maths so please forgive me if I am not as rigorous as you would like, but do not hesitate to correct me.

I want to find the equation of a torus (I mean the process, not just the final equation that I can find on Google). Knowing that a torus is the set of point on the circles having all their centers on another circle I came with something like:

A torus

Let $C_c$ be the "central" circle with a radius $R$ and and center $P_c(a, b, c)$. Also, let $M_1(x_1, y_1, z_1)$ be all the points on $C_c$. Let $C_a$ be an "auxiliary" circle (one that has $M_1$ as a center), $r$ his radius and $M_2(x_2, y_2, z_2)$ a point on that circle.

I'm looking for all the points $M_2$ to find the torus. Here's what I came to:

\begin{cases} (x_1 - a)^2 + (y_1 - b)^2 - R^2 = 0 \text{ (1)}\\ (x_2 - x_1)^2 + (y_2 - y_1)^2 - r² = 0 \text{ (2)} \\ \end{cases}

And I am stuck here, how can I transform those equations into a parametric form or a cartesian equation?



My goal is to find $x_2$ and $y_2$ here. So I decided to calculate $x_1$ and $y_1$ to use them in $(2)$.

From $(1)$ I get something like $x_1(x_1 - 2a) = -a^2 - y_1^2 - b^2 + 2by_1 + R^2$

But I am stuck here since I don't know how to "isolate" $x_1$

  • $\begingroup$ If you want to do anything with your torus $T$, say computing area, curvatures, or lengths of curves on $T$ you are better off with a parametrization than with an equation. Such an equation will be a fourth degree polynomial equation $p(x,y,z)=0$ from which you can't even decide without a lot of computation that it describes a torus. $\endgroup$ Commented Apr 10, 2014 at 17:46
  • $\begingroup$ You are right. The truth is I do not need to do anything with this torus right now, I ask this question for learning purposes only. $\endgroup$ Commented Apr 10, 2014 at 19:17

3 Answers 3


Implicit form

Add two more conditions to express the planes of your circles. Make sure your other equations are 3d as well. Then use e.g. resultants to eliminate $M_1$ and obtain a single implicit description of that torus.

For example, suppose $D(d,e,f)$ is the direction of the symmetry axis of the torus. Then your conditions can be written as

\begin{align*} \langle M_1-P_C,D\rangle &= 0 & (x_1-a)d+(y_1-b)e+(z_1-c)f &= 0 \\ \lVert M_1-P_C\rVert &= R & (x_1-a)^2 + (y_1-b)^2+(z_1-c)^2 &= R^2 \\ \langle M_2-M_1, (M_1-P_C)\times D\rangle &= 0 & (x_2-x_1)((c-z_1)e-(b-y_1)f)+\cdots&=0 \\ \lVert M_2-M_1\rVert &= r & (x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2 &= r^2 \end{align*}

Now you combine these three expressions, and in the process eliminate $x_1,y_1,z_1$. At least theoretically. Naively doing this using a resultant computation in sage takes longer than I'm willing to wait just now. Particularly since Wikipedia already has the quadric equation for a specific position, so all you have to do is apply translation and rotation to their formula. The default position is given as

$$(x^2+y^2+z^2 + R^2 - r^2)^2 = 4R^2(x^2+y^2)$$

Parametric form

For parametric, simply combine parametric descriptions of two circles. Start with $(r\cos\varphi,r\sin\varphi)$ then turn that into 3d, move it to the correct location and use $M_1$ as the center of the second circle.

  • $\begingroup$ Oh, you're right nothing tells $C_a$ is oriented the way that I want. $\endgroup$ Commented Apr 10, 2014 at 17:19
  • $\begingroup$ No problem for (2) and (4). I guess (correct me if I am wrong) that (1) tells that the vector "going from $P_c$ to $M_1$ should be orthogonal to $\vec{D}$ (which is normal since I do not want a deformed torus with "waves"). I guess (3) tells that $C_a$ should be on a plane parallel to $D$, but I do not understand how you make it work $\endgroup$ Commented Apr 10, 2014 at 19:44
  • $\begingroup$ If "$(M_1 - P_c) \times D$" is the cross product of the vector going from $P_c$ to $M_1$ and $\vec{D}$ then it makes sense and I understand. I am not comfortable with the notions you used. Anyway thanks, I know that I probably used the wrong vocabulary, and if so, do not hesitate to tell me. $\endgroup$ Commented Apr 10, 2014 at 20:00

In three dimensional space the equation of a circle is given by two equations: one for a sphere ($(x-a)^2+(y-b)^2+(z-c)^2=R^{2}$) and one for a plane ($\alpha x + \beta y + \gamma z = k$). Then your first locus has to be of this type.

Suppose we are smart and we choose the plane $z=0$ and $\left( 0,0,0 \right)$ as centre for $C_{c}$. Then the equations for the circle $C_{c}$ would be $x^{2} + y^{2} = R^{2}$ and $z=0$.

Then your intuition about how to find points on a torus is in the right direction, but you have to be careful. Once you fix a point in $C_{c}$, you want to draw a circle. But you are in three dimensional space, so you have to impose two conditions as above. What you forgot is to fix the plane "where your circle lives". Suppose you fix $\left(x_{0},y_{0},0\right) \in C_{c}$. Then the plane you have to consider is the one perpendicular to the tangent line to $C_{c}$ in $\left(x_{0},y_{0},0\right)$. Since we chose $C_{c}$ with nice coordinates, the plane will be $\left(y-y_{0}\right) = -\frac{x_{0}}{y_{0}} \left(x-x_{0}\right)$.

Now you can put equations together to find your locus.

  • $\begingroup$ Thanks! I would like to vote up but I can't since I only have a reputation of 6 here. I will try to apply your advices $\endgroup$ Commented Apr 10, 2014 at 17:21
  • $\begingroup$ You are welcome, and see you when you'll have reputation 15 :) $\endgroup$
    – Stefano
    Commented Apr 10, 2014 at 17:23
  • $\begingroup$ I've understood why you need to define a plan for $C_a$ but I do not understand how you end up with this plane equation $\endgroup$ Commented Apr 10, 2014 at 19:14
  • $\begingroup$ The equation of a sphere has squares on the terms on the left hand side. $\endgroup$ Commented Jun 16, 2016 at 7:14

You'd better use trigonometry.

Pick an angle $\theta$ to parametrize the "central" circle. Look for the parametric equations of the circle if you don´t know how.

Now pick another angle $\rho$ that will describe the "auxiliary" circles. Note that you must describe them in the same direction as the main radius $R$ is pointing. It is not as hard as it may seem.


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