492 Mr. A. Cayley on a Surface of the Fourth Order. 
four other foci, the points A, B, C, and a fourth point D lying 
in a circle with A, B, C, and which are such that, selecting any 
three at pleasure of the points A, B, C, D, the equation of the 
curve is in respect to such three points of the same form as it is 
in regard to the points A, B, C. 
Consider a given point M, on the principal section, then the 
equations 
gain Se! gel) eee meh FN 
BM CM’ CM AM’ AM’ BM 
belong respectively to three spheres: each of the spheres passes 
through the point M. The first of the spheres is such that, 
with respect to it, B and © are the images each of the other; 
that is, the centre of the sphere hes on the line BC, and the pro- 
duct of its distances from B and C is equal to the square of the 
radius; in like manner the second sphere is such that, with 
regard to it, C and A are the images each of the other; and the 
third sphere is such that, with regard to it, A and B are the 
images each of the other. The three spheres intersect in a 
circle through M at right angles to the principal plane (that is, 
the three spheres have a common circular section), and the equa- 
tions of this circle may be taken to be 
Oe ee 
AM” BM” CM’ 
It is clear that the circle of intersection lies wholly on the surface. 
The spheres meet the principal plane in three circles, which 
are the diametral circles of the spheres ; these circles are related 
to each other and to the points A, B, C, in hke manner as the 
spheres are to each other and to the same points. The circles 
have thus a common chord; that is, they meet in the point M 
and in another point M’. And MM! is the diameter of the circle, 
the intersection of the three spheres. 
It may be shown that M, M! are the images each of the other 
in respect to the circle through A, B, C. In fact, consider in 
the first place the two points A, B, and a circle such that, with 
respect to it, A, B are the images each of the other; take Ma 
point on this circle, and let Q be any point on the line at right 
angles to AB through its middle point, and join OM cutting the 
circle in M’; then it is easy to see that M, M’ are the images 
each of the other, in regard to the circle, centre O and radius 
OA (=OB). Hence starting with the pomts A, B, C and the ~ 
point M, let O be the centre of the circle through A, B, C, and 
take M! the image of M in respect to this circle ; then considering 
the circle which passes through M, and in respect to which B, C 
are images each of the other, this circle passes through M’; and 
