MR. R. LACHLAN ON SYSTEMS OF CIRCLES AND SPHERES. 
507 
So that we may write the above equation 
64 r 4 tan 3 o){A(l, 2, 3)} 4 — 
0, 
A(P, 2, 3). 
A(Q, 3, 1) 
, A(R, 
A(P, 2, 3), 
Pi.n 
Pi,2> 
Pi, 3 
A(Q. 3, 1), 
P:3,1’ 
P'S, 2’ 
p3,3 
A(R, 1, 2), 
P 3,U 
P'3,2’ 
p3,3 
• (58) 
Similarly we can find a> for the circles (P', Q, R), &c. 
40. If the given system (1, 2, 3) intersect at angles a, /3, y, equation (57) reduces 
easily to 
cos 2 w = sec s . cos (s— a). cos (s —/3 ). cos (s—y), .... (59) 
where 
2 s=a+/3+-y. 
Also for circle (P'QR) we shall find 
cos 2 w = cos s . sec (s — a). cos ( s—fi ). cos (s—y). 
41. It may be noticed that the system (P, Q, R, 4) is the orthogonal system to 
(1, 2, 3, S)—(see § 27). 
We have at once from equation (35) 
1 +—+=0. 
^P, 1 ^*Q, 2 7r B,3 
If S' be the circle through (P', Q', Pd), we have 
(60) 
L += L += L+_L =0 . 
7r P', 1 ^Q', 2 ^R', 3 7r S', 4 
But by (47), since S' is the inverse of S with respect to the circle (4), 
Hence we have 
"S, 4 *■$>, 4 7r 4,4 
1111112 
-+—-+-f--+— +-fi— — 0 ; 
7r P, 1 7r P', 1 ^Q, 2 7r Q', 2 ’Ll, 3 l7r K', 3 7f 4,4 
(61) 
or the sum of the reciprocals of the squares of the tangents, from the points of inter¬ 
section of three circles to the circles, is equal to the reciprocal of the square of the 
radius of the circle which cuts the circles orthogonally. 
3 T 2 
