276 Mr DAVIES on the Equations of Loci 



points. Then the equation of the circle is of the form 



cot (f)= tanX cos (6 K), .............................. (1.) 



And because a, B, and a,, /? are points in its course, we have also 

 cot a, = tan A. cos (ft, K), ........................ . ..... (2.) 



cota,,:= tan A cos (B u K), ............................. (3.) 



Divide (2.) by (3.), expand the cosines, and divide numerator and denomi- 

 nator of the right-hand number by cos K, which gives 



cot a, _ cos ft, + sin ft, tan K . 



cot a,, ~ cos ft,, + sin /? tan K' ' 



Or, resolving (4.) with respect to tan K, we have 



cot a,, cos ft, cot a, cos p*,, , 



"cot a, sin ft cot a, sin ft," 



cot a,, cos 8, cot a, cos B 

 sin/c I ......... (6.) 



V cot* a u 2 cot a,, cot a, cos (p a p,) + cot 2 a, 



cot a,, sin /3 cot a, sin ft, 



. - ...... 



V cot* a,, 2 cot a,, cot a, cos (ft, ft) + cot* a, 



either of which gives the longitude, or polar angle, of the centre of the circle. 



Again, from equations (2.) and (3.), we obtain 



cos- 1 ( cot a a cot \) cos- 1 ( cot a, cot X) = B u B t ; 

 or, taking cosine function of each side, and transposing, we obtain 



{(1 _ cot* a u cot 2 X) (1 cot 2 a, cot 2 A)P = cos (B a B) cot a, cot a,, cot 1 X ; 

 or, squaring and performing obvious reductions, 



cotX = 



. 

 V c ot* a a 2 cot a a cot a, cos (p,, p t )_+ cot 2 a., 



Inserting in (1.) the values of cot X, sin K, cos K, given by (6, 7, 8.) we find 

 ultimately the equation of the circle sought 



cot < = cosec (13 ' /?,) { cos 6 (cot a,, sin 6 cot a, sin /3J q= sin 6 (cot a^cos B, cot a, cos^,) } (& 



This may also be put under the form, and often advantageously, 

 , _ cot a, sin (B a :+: 0) cot a u sin (B. :+: 6) 



LUL *X/ ^~ . //O /O\ (ti 







