The two books of Apollonius Pergaeus, concerning tangencies, as they have been restored by Franciscus Vieta and Marinus Ghetaldus Apollonius Pergaeus John Lawson 868387657
<?xml version="1.0" encoding="UTF-8"?> <div> <p xmlns="http://www.w3.org/1999/xhtml" id="false"> greater of the given circles the line EH = the difference of the Radii of the given <br /> circles, and with A and B as Foci and EH Tranverſe Axis, let two oppoſite <br /> Hyperbolas be deſcribed KEK and LHL: then I ſay that KEK will be the <br /> locus of the centers of all the circles which can be drawn ſo as to be touched <br /> outwardly by both the given circles; and LHL will be the locus of the centers <br /> of all the circles which can be drawn ſo as to be touched inwardly by both the <br /> given circles. For taking any point K in the Hyperbola KEK, and drawing KA <br /> and KB, let theſe lines cut the given circumferences in F and G reſpectively: and make KQ = KA: then from the nature of the curve QB = EH, but by con- <br /> ſtruction EH = BG - AF, therefore QB = BG - AF and hence QG = AF, and <br /> then taking equals from equals KG = KF, which is a demonſtration of the 1ſt <br /> Caſe. </p> <p xmlns="http://www.w3.org/1999/xhtml"> <span class="caps" style="font-size: 75%">Then</span> with regard to the 2d Caſe, taking any point L in the Hyperbola <br /> LHL, and drawing LB and LA and producing them to meet the concave cir- <br /> cumferences in M and N, let alſo LR be taken equal to LB; then from the pro- <br /> perty of the curve AR = EH, but EH (by conſtruction) = BM - NA; there- <br /> fore AR = BM - NA, and NR = BM, and then adding equals to equals LN = <br /> LM, which is a demonſtration of the 2d Caſe. </p> <p xmlns="http://www.w3.org/1999/xhtml"> <span class="caps" style="font-size: 75%">Case</span> 3d. Suppoſe it be required that the circles to be deſcribed be touched <br /> outwardly by one of the given circles, and inwardly by the other. </p> <p xmlns="http://www.w3.org/1999/xhtml"> Then drawing AB, let it cut the convex circumferences in C and D, and the <br /> concave ones in P and O, biſect PD in E, and from E towards B ſet off EH = <br /> the ſum of the given radii. Then with A and B foci and EH tranſverſe axis, <br /> let two oppoſite hyperbolas be deſcribed KEK and LHL: and KEK will be the <br /> locus of the centers of the circles which are touched inwardly by the circle A <br /> and outwardly by the circle B; and LHL will be the locus of the centers of <br /> thoſe circles which are touched inwardly by B and outwardly by A. The demon- <br /> ſtration mutatis mutandis is the ſame as before. </p> <p xmlns="http://www.w3.org/1999/xhtml"> <span class="caps" style="font-size: 75%">Case</span> 4th. Suppoſe the given circle A to include B, and it be required that <br /> the circles to be deſcribed be touched outwardly by them both. </p> <p xmlns="http://www.w3.org/1999/xhtml"> Let AB cut the circumferences in C and D, P and O: and biſecting CD in <br /> I, and ſetting off from I towards P, IL = the ſum of the ſemidiameters of the <br /> given circles, and with A and B foci, and IL tranſverſe axis, deſcribing an <br /> ellipſe LKI, it will be the locus of the centers of the circles required. For <br /> taking any point K in the ellipſe, and drawing AK and BK, let AK be con- <br /> tinued to meet one of the given circumferences in G, and let BK meet the other <br /> F. Then from the property of the curve AK + BK = IL = AG + BF (by con- </p> </div>