Hogg. — On Orthogonal Circles. 587 



Let O be the pole of a conic S circumscribing the triangle ABC, and 

 let P, Q be the extremities of a chord of the conic passing through O. 

 If circles be drawn with their centres at P. Q and cutting the circle ABC 

 orthogonally, they will intersect each other on the circle whose centre is 

 at O and which cuts the circle ABC orthogonally. 



Let / ' , g , h 



S = J - + % + - = o ; 

 a (3 y 



and let X + fx. -+- v = o and A 1 = /* — v, fj} = v — A, v 1 = A — /x. We see 

 by inspection that the two points wiiose triiinear ratios are (j-, -, -J, 



(t, I,-) lie on S. The equation of the chord PQ is 

 VA 1 fj} v 1 ) 



Ut' + rfi + ^-o, 



which is satisfied by (/, g, h) — the co-ordinates of the pole of the conic S. 



The circles having their centres at P, Q and cutting the circle ABC 

 orthogonally will be 



a/X , bgY chZ 



A }jl v 



a/X i bgY , chZ 

 and for their intersections we have 





afX : bgY : chZ = — -. r :_ n fv : ^— r — tt- = XA 1 : uu 1 



J /av 1 ^v i/A 1 v x A A.//. 1 AV rr 



IT 1 



since /xv 1 — /aV = vA 1 — i^A = A/u 1 — A 1 /* = — 2 (A 2 ). 



Hence the locus of the intersection of Sj and S a is 



a/X + bgY + c/iZ = o, 



a circle having its centre at the pole (/, g, h) and cutting the circle ABC 

 orthogonally. 



If a circle cutting the circle ABC orthogonally meet the latter at the 

 extremities of a chord passing through the symmedian point of the 

 triangle ABC, it will pass through the two points whose pedal triangles 

 are equilateral. 



Any chord of the circle ABC which passes through the symmedian 

 point (a, b, c) of the triangle ABC may be written 



a b c 



where A + ^ -f- v = o. 



The co-ordinates of the pole of L with respect to the circle ABC 

 are given by 



a : fi : y = a\ : by. : cv ; 



hence the equation of the orthogonal circle is 



Aa 2 X + n&Y + vc % Z = o. 

 Its form shows that it passes through the pair of inverse points given by 



a 2 X = b 2 Y = c 2 Z. 



