[315 
315] 
WITH RESPECT TO THREE LINES AND A LINE. 
75 
double centre. And 
we obtain 
X y z a 
X+P + y + P + Z + P~ ' 
\!x + P 1 Vy + P + \/. z + P~°’ 
which are easily transformed into 
2 ); 
cc y \ ^ Ct 
x+P + y + P 1 Z + P 
yz t zx t xy 
0/ + P) (z + P) + (z +1) (X+P) + 0 + P) (y + P) ’ 
or, what is the same thing, 
6 (P + x) (P + y) (P + z) - x (P + y) (P + z) - y (P + z) (P + x) - z (P + x) (P + y) = 0, 
) $ -|- \, 0 + ¡Jb, 0 V \ 
9 (P + x)(P + y) (P + z) — yz (P + x) — zx (P + y) — xy (P + z) = 0, 
which give 
GP 3 + 5 P 2 (x + y + z) + 4 P (yz + zx + xy) + 2>xyz = 0, 
9P 3 + 9P 2 (x + y + z) + 8P (yz + zx + xy) + 6xyz = 0 ; 
locus of the double 
or, multiplying the first equation by 2, and subtracting the second, 
3P + (x + y + z) = 0; 
nonically in respect 
differently in the 
and we thus obtain for the locus of the single centre the equation 
x y z _ 
— 2 x + y z —2y + z + x —2 z + x + y 
antre. And we now 
or, what is the same thing, 
3 □ — 6a 2 
be 6a 2 5 
x? + y 3 + z s — (yz 2 + zx 2 + xy 2 + y 2 z + z 2 x + x 2 y) + 3 xyz — 0, 
which may also be written, 
— (—X + y + z) (x — y + z)(x + y — z) + xyz = 0. 
) 0 + 0 + fJL, 0 + V, 
The same result may also be obtained as follows: viz., observing that 
es P, a, b, c have 
: viz., writing the 
□ — 6 a 2 = b 2 + c 2 — 5a 2 = — 4«” — 25c, 
we have 
x — 3a 2 y — 35 2 £ — 3c 2 
P 2a 2 + 5c ’ P 25 2 + ca ’ P 2c 2 + «5 ’ 
and then by means of the equation 
a 2 5 2 c 2 I — ft 
2a 2 + 5c ' 25 2 + ac + 2c 2 + a5 
10—2