429 



| A D E A 2 I 

 | D B F B 3 m 



The intersection degenerates, if j E F C C 2 n 

 ! A 2 B 2 C 2 G p 

 I m n p o | 



In a non- centric surface, where V=0, we readily find that the 

 former of these eliminants has the same sign as (D 2 AB) ; and con- 

 sequently, that non-centric surfaces cannot have sections of opposite 

 species. It also appears, that to determine in a non-centric surface 

 the parabolic sections, we must take Imn such as to verify one of 

 the three eqq. 



A D E 

 DBF 



Imn 



A D E 



Imn 

 E F C 



/ m n 

 DBF 

 E F C 



=0. 



Problem. To determine the circular sections, when they exist. 



Result. Take the larger question, of ascertaining when two sur- 

 faces of the second degree intersect in a plane curve. Denote the 

 coefficients of the second surface by accents. Put a=Ap A'; 

 /3=Bp B' ; y=C P C' ; &c. and determine p by the eq. 



<t> 



y 



which involves p in the third degree. 



Then Imn will be determined (when the surds are real) by the 

 proportion 



To apply this to the problem of circular sections, it is only necessary 

 to suppose the second surface to be a sphere. 



The surface becomes one of Revolution, if (with oblique axes) 

 either system of three eqq. is fulfilled : 



f(l) o/3= 2 3 , ay=e 2 , $y=<f, 

 1(2) op = 2e, /3e=0?, y%= ej. 



If out of each triplet we eliminate p 2 and p, (for it seems easiest to 

 treat these as independent variables,) the result is two eqq. (expres- 

 sible by eliminants), which are the two general conditions for a sur- 

 face of revolution. 



2 K2 



