and Loci of Apollonius, S^c. 41 



Hence, the solution of this enunciated problem may be 

 made as follows : — In AOl and NN take P and R such that 

 QP.UR = m.n ; describe the circle AIB such that I being 

 an}' point in it, the angle lA right to B = ^ right ; through 

 P di'aw a line PG making the angle PG right to jNI =: RB 

 right to X, aiul describe the circle QOP through Q which 

 touches this line at P ; find the point O in this circle such 

 that PO : QO : : RB : RU ; through the point L in which 

 RB again cuts the circle AIB, draw AL to cut PO in H ; 

 describe the circle AOII ; through either point C in which 

 this circle cuts M]M, draw CA to cut the circle xVIB again in 

 I ; draw BI to cut i\IX in D. Then will AI and BI be 

 answeral)le lines. 



Moreover, it is evident that the angles PO and QO right 

 to M are respectively equals to the angles BR and UR right 

 to U and B, and that we can, therefore, determine PO and 

 QO without drawing PG or describing the circle QPOG. 



4. If in addition to the conditions of this third case, we 

 suppose the angle = zero, and B coincident v/ith A ; then, 

 it is evident, the problem becomes the "Section of Space" 

 of Apollonius. Here again it is evident RAH is a straight 

 line, and that the solution which I have given to this Apol- 

 lonian problem has been derived from the present more 

 general problem. 



5. We have PC.SD : QC.RD : : I : k, when the angle 9 

 has any finite magnitude ; and it is evident that when B = 

 zero, and that ]MM coincides wdth NX, we have PC.SC : 

 QC.RC : : / : A', which is the " Determinate Section" of 

 Apollonius. It is further evident that the solution just given 

 to this Apollonian problem has been derived from the present 

 more general one. 



REMARKS. 



If K be the other point in Avhich SB cuts the circle AIB ; 

 then it is obvious AK and QO intersect in a point T in the 

 circle AHO. 



If U be the other point of intersection of the circle OGPQ 

 witli any circle AOH, and that P^^ be drawn parallel to AL 

 to cut circle OGPQ in Y, then will VUA be in one straight 

 line. Therefore, as the point U is known independent of the 

 ratio, the limiting circles A^Jio'V passing through A and U 

 can be hence easily described. 



It is also evident QV is parallel to AK, &c. 



