1890.] connected with Bicircular Quartics. 91 



of the six points P t , P 2 ,...P 6 ; and then the circle touching the 

 curve at Q 1 and passing through P v i.e. P, must meet the curve 

 again in R the residual point of the seven consecutive points P. 



Otherwise, we may obtain R' the coresidual point of five 

 consecutive points P as in § 8, and then the circle which touches the 

 curve at P and passes through R' must meet the curve again in R. 



Incidentally we may notice that we have the theorem : if the 

 circle of curvature at any point P of a bicircular quartic meet the 

 curve again in Q, and if Q iy Q 2 , Q 3 , Q 4 be the four points of contact 

 of the bitangent circles at Q, then the four circles which can be 

 drawn touching the curve at these points respectively and passing 

 through P, cut the curve again in the same point R. 



Hence we see that R can only coincide with P, when Q 1 ,Q 2 ,Q 3 , Q 4 

 are the points of contact of the bitangent circles at P, in which 

 case Q must coincide with P, i.e. P must be a cyclic point. 



Thus at any point P on a bicircular quartic we can in general 

 draw other bicircular quartics having seven point contact at P, 

 and they will all cut the curve again in the point R. If however 

 P be a cyclic point these bicircular quartics will have eight-point 

 contact with the given quartic at P. 



11. All these theorems admit of translation so as to apply to 

 twisted quartics, we have merely to substitute in the enunciation 

 of any theorem the word plane for circle. In fact we have only 

 to prove that if any surface of the nth. order passes through 4n — 1 

 fixed points on the curve of intersection of two quadrics, it must 

 also pass through one other fixed point. 



Let U 2 , V 2 denote the two given quadrics, and U n any surface 

 of the nth degree passing through 4n — 1 fixed points on the curve 

 of intersection of U 2 , V 2 ; then we have to show that any other 

 surface of the nth. degree passing through these fixed points will cut 

 the curve of intersection of U 2 , V 2 in the same point as U n . Let 

 a surface U n _ 2 of the (n — 2)th order be drawn passing through 

 n (n — 2) arbitrary points on the curve of intersection of V 2 Avith 

 U n , and also through ^{n — l)(n — 2)(n — 3) arbitrary points on 

 the surface U n ; also let a surface V n _ 2 of the (n — 2)th order be 

 drawn passing through n (n — 2) arbitrary points on the curve of 

 intersection of U 2 with U n , and also through the same points on 

 U n as JJ„ ,. Then we have three surfaces U n , U .U ,V , . V, 

 each of the nth order and each passing through the same points, 

 Avhose number is 



4ra - 1 + 2n (n- 2) + %(n -l)(n- 2) (n - 3) = ±n (n 2 + Qn + 11) - i>, 

 which is two less than the number necessary to determine a surface 

 of the nth degree, any surface of the nth degree therefore which 

 passes through these points must be of the form 



u n + xu n _ 2 .u 2 + f ,v n _ 2 .v 2> 



VOI,. VII. PT. II. 8 



