586 Transactions. 



The locus of the centres of these circles is the conic whose equation is 



111 



+ + 5 = 0- 



o c c a a b 



This conic is the locus of points whose polars with respect to the triangle 

 formed by the tangents to the circle ABC at the vertices A, B, C pass 

 through the symmedian poiut of the ti-iangle ABC. 



To find the envelope of the circles cutting the circle ABC orthogon- 

 ally and having their centres on the curve whose equation is 



Consider the points whose trilinear ratios are 



(-, -i v)' (-» T' -)' where A + /* + v = o. 



\fX V A/ \V A fJi/ 



These points lie on the circle ABC, and the equations of the tangents to 

 the circle at them are 



-[A 2 +r V + L X l = 0. 



a b c 



a b c 



These lines will meet at the point 



a : (3 : y = a (A 4 - /*V 2 ) : b (^ - v 2 X 2 ) : c (v 4 - Ay 2 ), 

 which, since A 2 — /xv = /a 2 — vX = v l — \fx, reduces to 



a : p : y = a (A 2 + pv) : b (^ + vX) : c (/ + A/*). 



For the locus of the intersections of the tangents we have 



cab 



b c a 



giving . . _ __ a 2 _ £y . f? _ ya . y 2 _ a/3 



A • ** a? be' V ca' c 2 aV 



and hence the locus of the intersection of the tangents is 



The equation of the orthogonal circle may therefore be taken to be 



a 2 (A 2 + fiv) X + b 2 (/x 2 + vX) Y + c 2 (v 2 + A/*) Z = o ; 



or, writing — A. (it + v), — fi (v -f A), — v (A + //,) for A 2 , tr, v 2 respectively, 

 the equation of the circle is 



fj.v (- r/ 2 X + /rY + c-' 2 Z) -f rA (a 2 X - b 2 Y + c 2 Z) 

 + A/x (« 2 X + 6 2 Y - c 2 Z) = o, 



the envelope of which is 

 (-a 2 X + Z> 2 Y + c 2 Z)4 + (a 2 X - b 2 Y + c 2 Z)* + (a 2 X + // 2 Y - c 2 Z)i = o. 



