and Loci of Apollonitcs, ^c. 37 



right to R ; anrl therefore if L be the other point in which 

 RB cuts tlie circle Alli, and II that in -svhich OP cuts AL, 

 we have the angle OC right to P or H = BD or BI right to 

 R or L = AI or AC right to L or H; hence a circle can pass 

 through O, H, A, and C; but O, H^ and A are known points; 

 therefore the point C in Avhich the circle OIIA cuts MM is 

 known, and therefore the point I in which CA cuts the circle 

 AIB, and also the point D in which IB cuts NN. 



COMPOSITION. 



Through A and B describe the circle AIB, such that I 

 being any point in its circumference, the angle I A right to B 

 = right ; draw PG and QG, making the angle PG right 

 to M = RB right to N, and the angle QG right to M = SB 

 right to N ; describe the circle PGQ, and in it (on the same 

 or opposite side of PQ with G, according as q^ and j^ have 

 like or unlike signs) find O such that PO : QO : : /.RB : 

 ^.SB ; draw RB to cut the circle AIB in L; draw OP to cut 

 AL in II ; describe the circle OHA ; through either point C 

 in which the circle OHA cuts M^M draw CA to cut the circle 

 ABI in I ; draw BI to cut NN in D : then will AI and BI 

 be as required. 



For through the other point E in which the circle OCP 

 cuts GP, draw CE to cut QG in F. 



The angle EC right to P is = OC right to P or H = AC 

 or AI right to li or L = BI or BD right to L or R ; there- 

 fore, since the angle PC right to E = RD right to B, the 

 triangles PCE, RDB are similar, and CP or CO, right to E or 

 F is equal DR or DS right to B. 



And since the angle QG or QF right to P or C is equal 

 angle SB right to R or D, therefore the triangles CQF, DSB 

 are similar. Now from these two pair of triangles we 

 evidently have PE.SB : QF.RB :: PC.SD : QC.RD; but 

 (see Porism in Transactio7is for 1859) we have PE : QF : : 

 PO : QO : : /.RB : yt.SB, and therefore PE.SB : QF.RB 

 :: I : k', hence PC.SD : QC.RD : : / : k. 



DISCUSSION. 



Since there are two points of intersection C, there are two 

 solutions to the problem, both real or both imaginary, accord- 

 ing as these points arc real or imaginary. 



If the ratio -j^ be unrestricted as to sign, then it is evident 



