Manchester Memoirs, Vol.lix. (191 5), No. 13- n 



Then if O be taken as pole and OH as initial line, the 

 equation of the hyperbola is : 



(rcos 6 - a cos a) (rs'md- a sin a) = a 1 cos a sin a. 



Or 



/-' cos 6 sin 6 - ar sin 6 cos a - cz/- cos 6 sin a = 0. 

 Or 



- sin 26 = a sin (0 -f a). 

 2 



The intersections with the circle r—2a will satisfy the 



condition 



sin 2 = sin (0 + a). 

 Hence 



2 = 2 «7r + + a. 



Or 



2 = (2 n + i) 7T- (0 + a), 



where n is an integer or zero. The former solution gives 

 the point A for the position of P. The second solution 

 gives, for ?i = o, 1 or 2, three positions for P such that : 



3 3 3 3 3 



and these three points are the angular points of an equi- 

 lateral triangle. But it has been proved that if the angle 

 OLX (that is, ir — a) is three times the angle PON, the 

 angle A OB must be three times the angle POP; or in 

 other words, if 



0="-", 

 3 

 then 



_POB= 1 {LAOB). 



Therefore the two angles POB with the remaining two 

 positions of P, which are respectively 27r/3 and 4— 3 

 more than one-third of the angle AOB, must be equal to 

 one-third of the angles (AOB + 2ir) and (AOB + 4tt) 

 respectively. 



Crawford gives a general consideration of the inter- 

 section of a conic with a circle, from which may be 



