In geometry, Villarceau circles () are a pair of circles produced by cutting a torus obliquely through its center at a special angle. Given an arbitrary point on a torus, four circles can be drawn through it. One is in a plane parallel to the equatorial plane of the torus and another perpendicular to that plane (these are analogous to lines of latitude and longitude on the Earth). The other two are Villarceau circles. They are obtained as the intersection of the torus with a plane that passes through the center of the torus and touches it tangentially at two antipodal points. If one considers all these planes, one obtains two families of circles on the torus. Each of these families consists of disjoint circles that cover each point of the torus exactly once and thus forms a 1-dimensional foliation of the torus. The Villarceau circles are named after the French astronomer and mathematician Yvon Villarceau (1813–1883) who wrote about them in 1848.
Example Consider a horizontal torus in xyz space, centered at the origin and with major radius 5 and minor radius 3. That means that the torus is the locus of some vertical circles of radius three whose centers are on a circle of radius five in the horizontal xy plane. Points on this torus satisfy this equation:
0 = ( x 2 + y 2 + z 2 + 16 ) 2 − 100 ( x 2 + y 2 ) . {\displaystyle 0=(x^{2}+y^{2}+z^{2}+16)^{2}-100(x^{2}+y^{2}).\,\!}
Slicing with the z = 0 plane produces two concentric circles, x2 + y2 = 22 and x2 + y2 = 82, the outer and inner equator. Slicing with the x = 0 plane produces two side-by-side circles, (y − 5)2 + z2 = 32 and (y + 5)2 + z2 = 32. Two example Villarceau circles can be produced by slicing with the plane 3y = 4z. One is centered at (+3, 0, 0) and the other at (−3, 0, 0); both have radius five. They can be written in parametric form as
( x , y , z ) = ( + 3 + 5 cos ϑ , 4 sin ϑ , 3 sin ϑ ) {\displaystyle (x,y,z)=(+3+5\cos \vartheta ,4\sin \vartheta ,3\sin \vartheta )\,\!}
and
( x , y , z ) = ( − 3 + 5 cos ϑ , 4 sin ϑ , 3 sin ϑ ) {\displaystyle (x,y,z)=(-3+5\cos \vartheta ,4\sin \vartheta ,3\sin \vartheta )\,\!}
The slicing plane is chosen to be tangent to the torus at two points while passing through its center. It is tangent at (0, 16/5, 12/5) and at (0, -16/5, -12/5). The angle of slicing is uniquely determined by the dimensions of the chosen torus. Rotating any one such plane around the z-axis gives all of the Villarceau circles for that torus.
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 (see the figure to the right). Begin with a circle of radius r in the yz plane, centered at (0, R, 0):
0 = ( y − R ) 2 + z 2 − r 2 . {\displaystyle 0=(y-R)^{2}+z^{2}-r^{2}.}
Sweeping this circle around the z axis replaces y by (x2 + y2)1/2, and clearing the square root produces a quartic equation for the torus:
0 = ( x 2 + y 2 + z 2 + R 2 − r 2 ) 2 − 4 R 2 ( x 2 + y 2 ) . {\displaystyle 0=(x^{2}+y^{2}+z^{2}+R^{2}-r^{2})^{2}-4R^{2}(x^{2}+y^{2}).}
The cross-section of the swept surface in the yz plane now includes a second circle, with equation
0 = ( y + R ) 2 + z 2 − r 2 . {\displaystyle 0=(y+R)^{2}+z^{2}-r^{2}.}
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, on these tangent lines, z/y equals ±r/(R2 − r2)1/2, and choosing the plus sign produces the equation of a plane bitangent to the torus:
y r = z R 2 − r 2 . {\displaystyle yr=z{\sqrt {R^{2}-r^{2}}}.}
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