Prof. Cayley on Quartic Curves. 107 



points K, L and the opposite points K 7 , 1/ : then considering the 

 tangent circle as moving round A, A' until it returns to its ori- 

 ginal position, the points K, L, K', L' are always four distinct 

 points ; and K and some one (say L) of the two points L, 1/ will 

 describe the same oval, say the oval B ; while the opposite points 

 K ; , 1/ will describe the opposite oval B'. We have here the oval 

 A included in the oval B (and of course the opposite oval A' in- 

 cluded in the opposite oval B') . But the oval B, qua portion of 

 a spherical quartic, lies wholly in one hemisphere ; hence the two 

 ovals A, B lie wholly in one hemisphere. It is easy to see that 

 there is not in this case any other portion of the spherical quartic, 

 but that the two ovals A, B are the entire curve. 



Reverting to the case where we have in one hemisphere the 

 two ovals A, B external to each other, the spherical quartic may 

 comprise as part of itself another oval C. The ovals A and B, 

 qua ovals external to each other, have a common tangent circle (a 

 double tangent of the spherical quartic) which cannot meet the 

 oval C (for if it did we should have six points of intersection) ; 

 hence in the immediate neighbourhood thereof we have a circle 

 not meeting any one of the ovals A, B, C. We may consider 

 A, B, C as lying on the same side of this circle ; for if B were on 

 the opposite side to A, then B' would be on the same side ; and 

 so if (3 be on the opposite side, then C will be on the same 

 side ; that is, we have the three ovals A, B, C external to each 

 other, and in the same hemisphere. 



There may be a fourth oval, D, and it would be shown in a 

 similar manner that we have then the four ovals A, B, C, D ex- 

 ternal to each other and in the same hemisphere. But there 

 cannot be a fifth oval, E ; the proof is precisely the same as for 

 the theorem in piano; viz. taking within each of the five ovals a 

 point, and through these points drawing a conic, the conic would 

 meet each oval in two points, and therefore the plane quartic in 

 ten points, which is impossible. 



Passing from the sphere to the plane, the foregoing investiga- 

 tion shows that every plane quartic without nodes or cusps is 

 either a finite curve, or else the projection of a finite curve, of 

 one of the following forms : — 



1. a single oval. 



2. two ovals external to each other. 



3. two ovals, one inside the other. 



4. three ovals external to each other. 



5. 6. four ovals external to each other. 



The last case has been called (5, 6) for the sake of the follow- 

 ing subdivision, viz. : — 



5. the four ovals are so situate as to be intersected, each in 

 two points, by the same ellipse. 



