[128 
128] DEVELOPMENTS ON THE PORISM, &C. 139 
Theorem. In the case where the conics are replaced by the two circles 
x 2 + y 2 — R 2 = 0, (x — a) 2 + y 2 — r 2 = 0, 
then the discriminant, the square root of which gives the series 
A + Bg+Cl* + Df 3 + Ef 4 + &c., 
is 
(1 + f) [r 2 + £(r 2 + R 2 - a 2 ) + ?R 2 }. 
Write for a moment 
CIRCUM- 
A + B% + Cg 2 + Dg 3 + Eg 4 + &c. = V(1 + a£) (1 + bg) (1 + c£) > 
then 
A = l, 
2B = a + b + c, 
— 8C = a 2 + b 2 + c 2 — 2 be — 2 ca — 2 ab, 
16Z> = a 3 + b 3 + c 3 — a 2 (b + c) — b 2 (c + a) — c 2 (a + b) + 2abc, 
— 128 E = 5 a 4 + 5b 4 + 5c 4 — 4a 3 (b + c) — 4 b 3 (c+a) — 4 c s (u + 6) 
5.] 
+ 4a 2 6c + 46 2 ca + 4c 2 a& - 2b 2 c 2 - 2c 2 a? - 2a 2 b 2 , 
&c. 
given in my 
-circumscribed 
To adapt these to the case of the two circles, we have to write 
ows: 
r 2 (1 + af) (1 + (1 + c£> = (1 + {r 2 + %( r 2 +R 2 - a 2 ) + ?R% 
ciic U = 0 an 
3 development 
7 + V; viz. if 
and therefore 
c = 1, 
r 2 (a + b) = r 2 + R 2 — a 2 , 
r 2 ab = R 2 ; 
values which after some reductions give 
A = 1, 
r 2 .2B = 2 r 2 + R 2 - a 2 , 
- r 4 . 8(7 = (R 2 - a 2 ) 2 - 4<R 2 r 2 , 
r 6 . 15D = (R 2 - a 2 ) {(R 2 - a 2 ) 2 - 2r 2 (R 2 + a 2 )}, 
- r 3 . 128E = 5 (R 2 - a 2 ) 4 -8 (R 2 - a 2 ) 2 (R 2 + 2r 2 ) r 2 + 16aV. 
Hence also 
r 12 . 1024 (CE — D-) = {5 (R 2 - a 2 ) 4 -8 (R 2 - a 2 ) 2 (R 2 + 2r 2 ) r 2 + 16a 4 r 4 } {(R 2 - a 2 ) 2 - 4E 2 r 2 )} 
- 4 {(R 2 - a 2 ) 3 -2 (R 2 - a 2 ) (R 2 + a 2 ) r 2 } 2 , 
18—2