560 
SYSTEM OF EQUATIONS FOE, THREE CIRCLES 
[873 
In verification, observe that we have 
G — H =2tt 
-2A, 
(7' 
-H' = 
- ir + A, 
h 
CM 
II 
1 
Çb 
-A, 
H 
-F' 
= A, 
H-F = 
- 2B, 
H' 
-F' = 
7r + B, 
H — G' = 
-B, 
F 
-G' 
= 5, 
F - G = 
-2(7, 
F' 
-G’ = 
— 7T + C, 
F -H'= 
-G, 
G 
-H' 
= 0; 
hence 
(cos G — cos H) 2 + (sin G — sin H) 2 = 2 — 2 cos (G — H), = 2(1 — cos2X), = 4 sin 2 J., 
and we thus see that the sides are = 2sinX, 2 sin B, 2 sin (7 respectively. 
The first circle should pass through the points (cos G, sin G), (cos H, sin H); we 
ought therefore to have, for the first of these points, 
that is, 
1 + 2 ^ cos (G - F) + Sm -^— 
" sm 2 a 
sm a 
sin 2 A 
sin 2 a ’ 
sin (A — a) 
1 + 2 h ' cos 
sin a 
A , sin 2 {A — a) 
sin 2 a 
sin 2 A _ 
sin 2 a 
and, for the second of the points, the same equation. Write for a moment 
v sin A , sin(X-a) 
A = — , then = = X cos a — cos A ; 
sin a 
sm a. 
then the equation is 
1 + 2 (X cos a — cos A) cos A + (X cos a — cos A) 2 = X 2 , 
that is, 
1 — cos 2 A = X 2 sin 2 a, 
which is right. 
The second and third circles should intersect at the angle A', that is, we ought to 
have 
S ^~ cos ff ' - 8jn & ) - Bin <?' - sin H 
sin ¡3 
sm y 
sin 2 B sin 2 G sin B sin G ., 
+ I73U- + 2 „ ■ COS A , 
sm y 
sin 2 /3 sin 2 7 ^ sin A sin 7 
or, reducing and for cos (G'— H') substituting its value, =-cosX, the equation is 
si n»CB-£) sitf(g-Y) sin (j? (3) sin (C—y) _ sin» B sin» O „sin B sin C 
sin» ft sm» 7 sin ¡3 sin y sin» /3 + sin» 7 + 2 sin/9 sin 7 008 A ’ 
Writing here 
sin B _ sin G 
• /"A -* ) m 
sm /3 sm 7 
the equation is 
( F cos /3 — cos B) 2 + (Z cos 7 — cos (7) 2 + 2 ( F cos /8 — cos i?) (Z cos 7 — cos G) cos 
= F 2 + X 2 + 2YZ cos A',