Apollonius <Pergaeus>; Lawson, John, The two books of Apollonius Pergaeus, concerning tangencies, as they have been restored by Franciscus Vieta and Marinus Ghetaldus : with a supplement to which is now added, a second supplement, being Mons. Fermat's Treatise on spherical tangencies

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[Item 1.]
[2.] THE TWO BOOKS OF APOLLONIUS PERGÆUS, CONCERNING TANGENCIES, As they have been Reſtored by FRANCISCUSVIET A and MARINUSGHETALDUS. WITH A SUPPLEMENT.
[3.] THE SECOND EDITION. TO WHICH IS NOW ADDED, A SECOND SUPPLEMENT, BEING Monſ. FERMAT’S Treatiſe on Spherical Tangencies. LONDON: Printed by G. BIGG, Succeſſor to D. LEACH. And ſold by B. White, in Fleet-Street; L. Davis, in Holborne, J. Nourse, in the Strand; and T. Payne, near the Mews-Gate. MDCCLXXI.
[4.] PREFACE.
[5.] EXTRACT from PAPPUS’s Preſace to his Seventh Book in Dr. HALLEY’s Tranſlation. DE TACTIONIBUS II.
[6.] Synopsis of the PROBLEMS.
[7.] PROBLEMS CONCERNING TANGENCIES. PROBLEM I.
[8.] PROBLEM II.
[9.] PROBLEM III.
[10.] The GENERAL Solution.
[11.] PROBLEM IV.
[12.] PROBLEM V.
[13.] The general Solution.
[14.] PROBLEM VI.
[15.] The general Solution.
[16.] PROBLEM VII.
[17.] LEMMA I.
[18.] PROBLEM VIII.
[19.] Mr. Simpſon conſtructs the Problem thus.
[20.] PROBLEM IX.
[21.] LEMMA II.
[22.] LEMMA III.
[23.] PROBLEM X.
[24.] PROBLEM XI.
[25.] PROBLEM XII .
[26.] LEMMA IV.
[27.] LEMMA V.
[28.] PROBLEM XIII.
[29.] PROBLEM XIV.
[30.] SUPPLEMENT. PROBLEM I.
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78
PROBLEMS
CONCERNING
DETERMINATE SECTION.
PROBLEM I.
TO cut a given indefinite right line in one point, ſo that of the ſegments
intercepted between that point and two other points given in the inde-
finite right line, the ſquare of one of them may be to the rectangle under the
other and a given external right line, in a given ratio.
In the given indefinite right line let be aſſigned the points A and E, it is then
required to cut it in the point O, ſo that AO2 may be to OE into a given
line AU in the ratio of R to S;
which ratio let be expreſſed by AI to AU,
ſetting off AI from A either way, either towards E or the contrary;
and
then from A and I erect two perpendiculars AY equal to AE, and IR
equal to AI, and theſe on the ſame ſide of the given indefinite line, if AI
was ſet off towards E;
but on oppoſite ſides, if AI was ſet off the other way.
The former conſtruction I will beg leave to call Homotactical, and the latter
Antitactical.
Let now the extremities of theſe perpendiculars Y and R be
joined, and upon YR as a diameter let a circle be deſcribed, I ſay that the
interſection of this circle with the given indefinite line ſolves the Problem.

If it interſects the line in two places, the Problem admits of two Solutions;

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