traced upon the Surface of the Sphere. 91 



As, from any point on the sphere, we may proceed in four different di- 

 rections so as to cut the given circle under a given angle, still so related 

 that one pair will form adjacent arcs of one great circle, and the other two 

 of another, so there will be two great circles which fulfil the condition 

 required. Each of these has two centres diametrically opposite on the 

 sphere ; and the values of sin K given by extracting the root of (14), and 

 the values of COSK from (15) refer to the longitudes of these four centres, 

 and determine them. The values of sin A and cos A can now be determined 

 from (6, 7), and the question is completely solved. 



When the given angle is a right one, cos e / =0 / , and hence A=B C 0, 

 and the radical vanishes, as well as certain other of the terms involved in 

 cos K and sin K. We thus get 



. ,E'E 2 -D* 2E 2 -D 2 E 



_ (16.) 



D z ( D 2 E 2 ) + 2 D 2 E 2 D 2 



_ _ _ ' ' _ _ 



4 D 2 E 2 + (E 1 D') 2 ~E 2 + 



agreeing with the result obtained in (IX, 7, 8), where this particular case 

 was the subject of examination. 



XIII. 



To find the equation of a circle through three given points on the sphere. 



Let the circle be denoted by 



cos> = cos A cos (p + sin A sin (p cos (6 /c) n.) 



and the three points be a/3, a,/?,, and a u p if Then we have, because (1) 

 passes through these three points, to find $, K, A, from the three following 

 equations : 



cosp = cos A cos a + sin A sin a cos (/? k) (2.) 



cosp = cos A cos a, + sin A sin a, cos (/3 y K) (3.) 



cos p = cos A cos a u + sin A sin a,, cos(/3,, K) (4.) 



