569 
875] WHICH CUT EACH OTHER AT GIVEN ANGLES. 
or, completing the reduction by the substitution of the value of 4/3 2 , this is 
y {(Q 2 sin 2 G + R 2 sin 2 B) + 2QR (cos A + cos B cos G)} 
+ QB {Q (cos B + cos C cos A) + R (cos G + cos A cos B)} 
= ± 4(3QR a/{— (1 — cos 2 A — cos 2 B — cos 2 (7-2 cos A cos B cos G)), 
viz. we have thus two values of the radius y (= P); and to each of these there 
corresponds a single value of the abscissa x, given by 
4/3x = R 2 — Q 2 + 2 (R cos B — Q cos G) y. 
The two values become equal, if A+B±G = 7r; in this case the three circles 
meet in a pair of points (x 1? y^, (x 1? — yi). In fact, writing A + B + G — 7r, and thence 
cos A = — cos (B + G), ’= — cos B cos G + sin B sin G, &c., 
we find 
[Q 2 sin 2 C + 2QB (cos A + cos B cos G) + B 2 sin 2 B\ y 
+ QR {Q (cos B + cos (7cos A)+ R(cos (7+ cos A cos B)} = 0, 
(Q sin G + B sin Bf y + QR (Q sin C + B sin B) sin A =0, 
that is, 
or, throwing out the factor Q sin G + R sin B, this is 
(Q sinC + R sin B) y + QR sin A = 0, 
and we then have 
4/3x = R 2 - Q 2 - 2 (R cos B - Q cos G) p „ 
x iismn+ysmG 
— t, • n 1 r\——n K-K sin B + Q sin G) (R 2 — Q 2 ) — 2 (R cos B — Q cos G) QR sin J.}. 
R sm B + Q sm G y v ' v ^ 1 ^ J 
The term in 
which is 
or, finally 
is here 
B? ( sin B) 
+ R 2 Q ( sin G — 2 sin A cos B) 
+ RQ 2 (— sin B + 2 sin A cos G) 
+ Q 3 (- sin G), 
= R 3 ( sin B) 
+ R 2 Q (— sin (7 + 2 sin B cos A) 
+ RQ 2 ( sin B — 2 sin G cos A) 
+ Q 3 (- sin G) 
= (R 2 + Q 2 + 2RQ cos A) (jR sin B — Q sin (7), 
= 4/3 2 (R sin B — Q sin G), 
— QR sin A 
y R sin B + Q sin G ’ 
¡3 (R sin B — Q sin G) 
R sin B + Q sin (7 
C. XII. 
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