284 Mr DAVIES on the Equations of Loci 



,_ _ (cosp ; sinA <) cos/c /) cospaSinA^s/Qcosfl+fcosp^sinA^sinK,, cos p,, sin A, sin K ; ) sin 

 ^ " cos p, cos A a cos g a cos A, 



cos p sin A sin /c., cos p,, sin A. sin K, 



Hence, tan/c = ' . " *" . ' L .. . (8.) 



cos p, sm A,, cos K,, cosp a sin A y cos K, 



The radius to which this is referred, is 

 = {cos* p, sin 2 A,, 2 cos p, cos p x/ sin A, sin \ cos K U K, + cos 2 p,, sin 2 A,} 2 (9.) 



And tan* _ -j-Icos 5 ^, sin* A, -2cos p, cosp /; sin \ sin A, cos K a -K, +cos*p /< sin* A'} 2 , 1Q . 



cos p y cos A /; cos o a cos A y 



The circle is hence completely determined. 





 X. 



To find the equation of a great circle which passes through a given 

 point a, /5 7 , and is perpendicular to a given great circle, j 



Let cot (p = tan A, cos (0 K) (1.) 



be the given circle ; and let that sought be denoted by 



cot(f> = tan A cos (6 k) (2.) 



But this passes through a, (3, , and therefore, also, 



cota, = tanAcosd^ K) (3.) 



Also the inclination of two circles being 



cos e, = cos A cos A, + sin A sin A, cos (x 1 K) (4.) 



7T 



but in the present case they are perpendicular, e 7 = g, and hence (4) be- 

 comes 



cotA = tanA.cos/c K, ............................. (5.) 



Between (3) and (5) we have to determine K and A. 

 We have, by division, 



tan\ = cg= = cosft + sin 



tan a, cos K, K cos K, + sin K, tan K 



tan/c = tan a ' cos & ~ tan X ' cos K ' (6.) 



tan a, sin /3 7 tan \ sin K t 



