Hogg. — Isogonal Transformations. 335 



the angle at which it cuts that circle. Then if p < R, the 

 radius of the circle ABC, L will isogonally transform into the 

 hyperbola S = //3y + mya + nafi = o, whose eccentricity (e) is 



given by the relation e = sec ^. From this we may deduce the 



following expression for the eccentricity : — 



2R 



e 2 



P + R 



We now proceed to the case where p > R. Suppose the 

 line L' = l'a + ??i'/3 + n'y — o be drawn parallel to L and 

 passing through the pole of that line with respect to the circle 

 ABC. Let p' be the distance of L' from the centre of the 

 circle ABC, and let L' cut that circle at the angle <£'. 



The line L transforms into an ellipse, and the angle (i//) 

 between its equi-conjugate diameters, expressed in terms of 

 the invariants ©, ©', is given by 



©' 2 - 4 © 



COS- \b = — 



© 2 



Taking S to be 2//3y + 2mya + 2rca/3 = o, we have 



©' = — 2 (/. cos A + m cos B + n cos C) 



© = - P sin 2 A - m a sin 2 B - n % sin 2 C 



+ 2 mn sin B sin C + 2 nl sin C sin A 

 4- 2 Im sin A sin B, 



n 



whence cos 2 \b = zz— -—— — zr— — — — 



(I cos A + m cos B + n cos C) 2 



where 



Q = I 2 -j- m 2 + 7i 2 — %nn cos A — 2nl cos B — 11m cos C. 



We also have 



2 R 2 (I cos A + m cos B + n cos C) 2 



therefore 



and pp' = R 2 , 



COS lb = — = L — = COS lb 



p R 



Hence we derive the result that the angle between the equi- 

 conjugate diameters of S is equal to the angle at which L' cuts 

 the circle ABC. 



Moreover, since L and L' are parallel, their isogonal trans- 

 formations, S and S', will intersect in four concyclic points : 

 the chords of the circle joining these four points will be equally 

 inclined to the axes of each of the conies : in other words, the 

 axes of the two conies will be parallel, and therefore the 

 Simson lines of the two points in which L' cuts the circle ABC 



