Villarceau Circles - Existence and Equations

Existence and Equations

A proof of the circles’ existence can be constructed from the fact that the slicing plane is tangent to the torus at two points. One characterization of a torus is that it is a surface of revolution. Without loss of generality, choose a coordinate system so that the axis of revolution is the z axis. Begin with a circle of radius r in the xz plane, centered at (R, 0, 0).

Sweeping replaces x by (x2 + y2)1/2, and clearing the square root produces a quartic equation.

The cross-section of the swept surface in the xz plane now includes a second circle.

This pair of circles has two common internal tangent lines, with slope at the origin found from the right triangle with hypotenuse R and opposite side r (which has its right angle at the point of tangency). Thus z/x equals ±r / (R2 − r2)1/2, and choosing the plus sign produces the equation of a plane bitangent to the torus.

By symmetry, rotations of this plane around the z axis give all the bitangent planes through the center. (There are also horizontal planes tangent to the top and bottom of the torus, each of which gives a “double circle”, but not Villarceau circles.)

We can calculate the intersection of the plane(s) with the torus analytically, and thus show that the result is a symmetric pair of circles, one of which is a circle of radius R centered at

A treatment along these lines can be found in Coxeter (1969).

A more abstract — and more flexible — approach was described by Hirsch (2002), using algebraic geometry in a projective setting. In the homogeneous quartic equation for the torus,

setting w to zero gives the intersection with the “plane at infinity”, and reduces the equation to

This intersection is a double point, in fact a double point counted twice. Furthermore, it is included in every bitangent plane. The two points of tangency are also double points. Thus the intersection curve, which theory says must be a quartic, contains four double points. But we also know that a quartic with more than three double points must factor (it cannot be irreducible), and by symmetry the factors must be two congruent conics. Hirsch extends this argument to any surface of revolution generated by a conic, and shows that intersection with a bitangent plane must produce two conics of the same type as the generator when the intersection curve is real.