I02 KANSAS UNIVERSITY QUARTERLY. 



2a(b-|-ak) — 2ab — a-k 

 /3'- . 



2a— 2a-|-2bk 



Froii] this is seen that the centers of corresponding orthogonal 

 circles in the two pencils form projective point-ranges. The two 

 pencils are, therefore, also projective and their prodnct is a bi-cir- 

 cular curve of the fourth order which degenerates into two circles. 

 To obtain the equation of these circles we have to eliminate A and 

 A' from the following equations: 



x2-)-y2 — 2Ax = o (I) 



(x— a)2 + (y— b)2-~2A'(x— yk— a + bk)=o (II) 



2Aa— a2— b2 



A'= . (Ill) 



2a — 2A— 2bk 



From (I) follows 



hence 



A': 



x~+y- 



a(x3J-y2)_x(a2-fb2) 



2x(a — bk) — (x2+y2) 

 Substituting this value in II, there is 



r(x— a)2 + (y— b)-l r2x(a-bk)— (x^ + ys)"]— 



— 2 X — yk + bk — a a(x-+y2) — x(a2-^b2) =o. 



After some transformations and reductions this equation may be 

 written in the conspicuous form 



rx2+y2~x(a— bk— b,/i+k"2)— y(b + ak-fal i+k2)1x 



(7) 

 XJ x2+y2_x(a— bk + bl i -l k ^ ) _y ( b + ak— a 1 ' i + k ^ ) =0. 



This is the equation of tlie product of the two projective special 

 pencils of circles and, evidently, represents two circles 



x^H-y-— x(a— bk— bl i+k>^)— y(b + ak + al ' i + k2)=:o, (8) 



x2-fy2_x(a— bk + bl i+k^)— y(b+ak--al/ i+k2)=o, (9) 



which both pass through the origin and through the point (a, b), 

 the two finite double points. 



