142 UNIVERSITY OF COLORADO STUDIES 



or 



a 2 — b 2 — 2abk 



& 



2a — 2a — 2bk 

 2a (b + ak) — 2ab — a 2 k 



2a — 2a + zbk 



From this is seen that the centers of corresponding orthogonal circles 



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



projective and their product is a bi-circular curve of the fourth order. 



To obtain the equation of this curve we have to eliminate X and A. ' from 



the following equations : 



X 2 _j_ y2 _ 2 x x = (I) 



(x — a) 2 + (y — b) 2 — 2 V (x — yk — a + bk) = o (II) 



X / = g^g-g'-^' (m) 



2a — 2 X — 2bk 



From (I) follows 



x 2 + y 2 

 \ = — , 



2X 



hence 



f a (x 2 + y 2 ) —x(a 2j r b 2 ) 

 2x(a — bk) — (x 2 -f- y 2 ) 

 Substituting this value in II, there is 



[(x — a) 2 + (y — b) 2 l [2X (a — bk) — (x 2 + v 2 )l — 



2 [x — yk + bk — a\ la (x 2 + y 2 ) — x {a 2 + b 2 ) 1 = o. 

 This equation may also be written in the conspicuous form 

 V X 2 _j_ y 2 _ x ( a _ bk —bv'i + k 2 ) —y{b + ak + aV \ + k 2 ) j X 



(22) 



Xx 2 +y 2 —x{a — bk + b\ 1 + £ 2 ) — v(6 + a& — al 7 i + & 2 )~|=o 



and represents two circles 



^2 _^y2_ x ( a _i ) k_i )V / Y+T 2 ) —y(b + ak + aV 1 + k 2 )=o, (23) 



^4-^— jc(a — &£-+- Ji/ 1 + k 2 ) —y(b-\- ak + aVi + & 2 )=o, (24) 



