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"> equal to UE: but RM is given in magnitude, being the radius of the given <br /> ſphere, therefore UE is alſo given in magnitude. And ſince OE is perpen- <br /> dicular to the plane DE, it will be alſo to the plane PU which is parallel <br /> thereto. UE then being given in magnitude, and being the interval be- <br /> tween two parallel planes DE, PU, whereof DE is given in poſition by hypo- <br /> theſis, the other PU will alſo be given in poſition. In the ſame manner it <br /> may be proved that the planes GH, IN, are given in poſition, and that the <br /> lines OG, OI, are perpendicular thereto reſpectively, and each alſo equal to <br /> OM. A ſphere therefore deſcribed with center O and OM radius will touch <br /> the three planes PU, GH, IN, given in poſition: but the point M is given, <br /> being the center of the given ſphere. The queſtion is then reduced to this, <br /> Having three planes given PU, GH, IN, and a point M, to find the radius <br /> of a ſphere which ſhall touch the given planes, and paſs through the given <br /> point; which is the ſame as the preceeding Problem. [And this radius be- <br /> ing increaſed or diminiſhed by MR, according as R is taken in the further or <br /> nearer ſurface of the given ſphere, will give the radius of a ſphere which will <br /> touch the three given planes DE, DB, BC, and likewiſe the given <br /> ſphere. ] </p> <p xmlns="http://www.w3.org/1999/xhtml"> <span class="caps" style="font-size: 75%">By</span> a like method, when among the Data there are no points, but only <br /> planes and ſpheres, we ſhall always be able to ſubſtitute a given point in the <br /> place of a given ſphere. </p> <h2 xmlns="http://www.w3.org/1999/xhtml"> <span class="headingNumber">43.</span> <span class="head">PROBLEM VII.</span> </h2> <p xmlns="http://www.w3.org/1999/xhtml"> <span class="caps" style="font-size: 75%">Having</span> two points H, M, as alſo two planes AB, BC, given, to find a <br /> ſphere which ſhall paſs through the given points, and touch the given <br /> planes. </p> <p xmlns="http://www.w3.org/1999/xhtml"> <span class="caps" style="font-size: 75%">Draw</span> HM and biſect it in I, the point I will be given, through the <br /> point I let a plane be erected perpendicular to the right line HM, this plane <br /> will be given in poſition, and the center of the ſphere required will be in this <br /> plane. But becauſe it is alſo to touch the planes AB, BC, its center will be <br /> alſo in another plane given in poſition (by what has been proved, Prob. IV.) and therefore in a right line which is their interſection, given in poſition, <br /> which let be GE; to which line GE from one of the given points M demit- <br /> ting a perpendicular MF, it will be given in magnitude and poſition, and <br /> being continued to D ſo that FD equals MF, the point D will be given; and, from what has been proved before, will be in the ſpherical ſurface. </p> </div>