TOL. VIII.] PHILOSOPHICAL TRANSACTIONS. Ill 



as yet used, are contained in one and the same general analysis. Now to show 

 this, let there be given the circle, whose centre is A, (fig. 8), also the points H 

 and I ; and let there be required the point K, to which, from the points H and 

 I, let there be drawn HK, IK, and the tangent KD. Then, from A draw any 

 line AG, meeting HK in E, IK in B, and the tangent KD in D, producing 

 such of the lines as are necessary. This done, it is plain, because of the equal 

 angles EKD, DKB, and the right angle AKD, that the three lines AE, BE, 

 DE, will be always harmonical proportionals. Therefore drawing KC, IF, 

 H G perpendicular to AE, and calling 



AK, q then by the method I formerly used in the second analysis of this 

 AC, a problem, we have this general equation, ndaa — bzaa — nqqa 4- 

 CK, e hqaa = ndee — zhee + ibnae + 2zdae — dqqe — zqqe. Sup- 

 HG, b pose now AG to be perpendicular to HI, there will be no variation 

 AG, d in the equation, except that AF and AG, that is d and z, will be 

 FA, z equal. Putting therefore d for z, it makes ndaa — bdaa — nqqa -\- 

 FI, ??, bqqa = ndee — dbee -\- ibnae + iddae — idqqe; or dividing 



,, , , 7, ^, qqa . 2bnae + Iddae — 2dqqe i • i . , 



all by ?za — db, then a a ~- = ee -{ d — db ^' which is the 



very same that I deduced from my first analysis, though by a different way, and 

 which I lately sent you, constructed after an easy manner. 



Next suppose AG to coincide with AH; then HG or Z' will become nothing. 

 Expunging therefore out of the equation all the terms in which b is found, there 

 will remain ndaa — nqqa = ndee -{- 2zdae — dqqe — qqze. And this, you 

 may recollect, I gave as my second thoughts, and another similar to it, for the 

 case in which AG passes through I. 



Again, let us suppose AG to bisect the angle HAI; then, because of the 

 similar triangles HAG, lAF, it will be, as HG to GA, so IF to FA, or as 

 b'.d'.:n:z, and nd =. bz. Then, taking away equals, it is bqqa — nqqa = 

 2bnae + Izdae — dqqe — qqze : the very same, as I perceive by your letter, 

 which M. Huygens has constructed. 



Lastly, suppose HG to bisect HI ; therefore HG = IG, or b =: n; and, by 



taking away equals, it is bdaa — bzaa = bdee — bzee -{- 2bbae -}- Izdae 



dqqe — qqze; which, though not very difficult, none of us has yet constructed. 

 But these, as well as the general equation itself, may be divided into two others, 

 by putting, as you know, for aa ov ee, its value qq — ee or qq — aa. 



You see then, whatever has hitherto been done, is resolved into the same 

 analysis ; which comprehends also infinite other constructions by the given 

 circle and an hyperbola. But it is of no great consequence to investigate them 

 since in this problem, though formerly solutions were wanting, yet now there 



