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Ellcrx W. Davis 



the area of P 2 'P 1 'P 5 'P 1 is i times that of P 2 'P^P Z 'P^' which is 

 l{(.v ? ;— x./)}^" — .\\"(y s ' — y./)}. Thus the sum of the three 

 areas of the type is 



It can be verified geometrically that in case P ± , P 2 , and P 3 all 

 belong to the same circular ray both the real and the imaginary 

 parts of the triangle area vanish. By parallel projection it then 

 follows that the area vanishes when the three vertices all belong 

 to the same elliptic line. 



The Circle 

 We consider first 



with its circle of black points x'- -\- y'~ = I. 

 Suppose that x' > i and x" = o, then 



y' = o and y" = i V ( x' 2 — I ) . 



Thus, as the black point of a red vector belonging to the circle 

 describes the :r-axes, the blue point describes a blue rectangular 

 hyperbola, a supplementary curve to the circle. 1 Moreover, since 

 a revolution of the axes, keeping them rectangular, does not alter 

 the equation of the circle, there is such an hyperbola tangent to 

 every point of the black circle and coincident with a revolved 

 position of this hyperbola. The black circle separates the part of 

 the plane where elements of the circle exist from the part where 

 they do not. 



O — a) 2 + (v — b)- = r 2 , 



differs from the case just considered only by a magnification by 

 r and a shift of center to {a, b). 



As Poncelet 2 long ago showed, they all pass through / and /, 



1 Poncelet, Traite des Proprietes Projectives, p. 29. 



2 Poncelet, Traite, p. 48. 



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