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

Table of contents

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[31.] PROBLEM II.
[32.] PROBLEM III.
[33.] PROBLEM IV.
[34.] PROBLEM V.
[35.] PROBLEM VI.
[36.] General Solution.
[37.] A SECOND SUPPLEMENT, BEING Monſ. DE FERMAT’S Treatiſe on Spherical Tangencies. PROBLEM I.
[38.] PROBLEM II.
[39.] PROBLEM III.
[40.] PROBLEM IV.
[41.] PROBLEM V.
[42.] PROBLEM VI.
[43.] PROBLEM VII.
[44.] LEMMA I.
[45.] LEMMA II.
[46.] LEMMA III.
[47.] LEMMA IV.
[48.] LEMMA V.
[49.] PROBLEM VIII.
[50.] PROBLEM IX.
[51.] PROBLEM X.
[52.] PROBLEM XI.
[53.] PROBLEM XII.
[54.] PROBLEM XIII.
[55.] PROBLEM XIV.
[56.] PROBLEM XV.
[57.] Synopſis of the PROBLEMS.
[58.] THE TWO BOOKS OF APOLLONIUS PERGÆUS, CONCERNING DETERMINATE SECTION, As they have been Reſtored by WILLEBRORDUS SNELLIUS. By JOHN LAWSON, B. D. Rector of Swanſcombe, Kent. TO WHICH ARE ADDED, THE SAME TWO BOOKS, BY WILLIAM WALES, BEING AN ENTIRE NEW WORK. 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. MDCC LXXII.
[59.] ADVERTISEMENT.
[60.] EXTRACT from PAPPUS's Preface to his Seventh Book in Dr. HALLEY's Tranſlation. DE SECTIONE DETERMINATA II.
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            <s xml:id="echoid-s704" xml:space="preserve">
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            diameter of the circle given in poſition, and therefore the points A and D
              <lb/>
            will alſo be given.</s>
            <s xml:id="echoid-s705" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s706" xml:space="preserve">
              <emph style="sc">Let</emph>
            us now ſuppoſe the thing done, and that the center of the ſphere
              <lb/>
            ſought is E, which, as obſerved before, muſt be in the line CB. </s>
            <s xml:id="echoid-s707" xml:space="preserve">Drawing
              <lb/>
            EF, EA, ED, theſe lines muſt be equal, ſince the points F, A, D, have
              <lb/>
            been ſhewn to be in the ſurface of the ſphere. </s>
            <s xml:id="echoid-s708" xml:space="preserve">But theſe lines EF, EA, ED,
              <lb/>
            are in the ſame plane, ſince FB and AD are parallel, and BC perpendicular
              <lb/>
            to each of them. </s>
            <s xml:id="echoid-s709" xml:space="preserve">If therefore a circle be deſcribed to paſs through the three
              <lb/>
            points F, A, D, whoſe center is E, it will be in the line CB, and will be
              <lb/>
            the center of the ſphere required.</s>
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          </p>
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        <div xml:id="echoid-div38" type="section" level="1" n="38">
          <head xml:id="echoid-head45" xml:space="preserve">PROBLEM II.</head>
          <p>
            <s xml:id="echoid-s711" xml:space="preserve">
              <emph style="sc">Having</emph>
            three points N, O, M, given, and a plane AD, to deſcribe a-
              <lb/>
            ſphere which ſhall paſs through the three given points; </s>
            <s xml:id="echoid-s712" xml:space="preserve">and alſo touch the
              <lb/>
            given plane.</s>
            <s xml:id="echoid-s713" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s714" xml:space="preserve">
              <emph style="sc">Let</emph>
            a circle ENOM be deſcribed to paſs through the three given points,
              <lb/>
            it will be in the ſurface of the ſphere ſought, from what was ſaid under the
              <lb/>
            former Problem. </s>
            <s xml:id="echoid-s715" xml:space="preserve">From it’s center I let a perpendicular to it’s plane IA be
              <lb/>
            erected; </s>
            <s xml:id="echoid-s716" xml:space="preserve">the center of the ſphere ſought will be in this line IA; </s>
            <s xml:id="echoid-s717" xml:space="preserve">let the line
              <lb/>
            IA meet the given plane in the point A, which point will be therefore given.
              <lb/>
            </s>
            <s xml:id="echoid-s718" xml:space="preserve">From the center of the given circle ENOM, let a perpendicular to the given
              <lb/>
            plane ID be drawn, the point D will then be given, and therefore the line
              <lb/>
            AD both in poſition and magnitude, as likewiſe the lines ID, IA, and the
              <lb/>
            plane of the triangle ADI. </s>
            <s xml:id="echoid-s719" xml:space="preserve">But the plane of the circle NOM is alſo given
              <lb/>
            in poſition, and therefore alſo their common ſection EIF, and hence the
              <lb/>
            points E and F.</s>
            <s xml:id="echoid-s720" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s721" xml:space="preserve">Suppoſe now the thing done, and that the center of the ſphere ſought is B.
              <lb/>
            </s>
            <s xml:id="echoid-s722" xml:space="preserve">Draw BE, BF, and BC parallel to ID. </s>
            <s xml:id="echoid-s723" xml:space="preserve">Since the triangle ADI, and the
              <lb/>
            line EIF are in the ſame plane, therefore BE, BF, BC, will be in the ſame
              <lb/>
            plane. </s>
            <s xml:id="echoid-s724" xml:space="preserve">But the line ID is perpendicular to the given plane, therefore the
              <lb/>
            line BC parallel to it, will alſo be perpendicular to the given plane. </s>
            <s xml:id="echoid-s725" xml:space="preserve">Since
              <lb/>
            then a ſphere is to be deſcribed to touch the plane AD, a perpendicular BC
              <lb/>
            from it’s center B will give the point of contact C; </s>
            <s xml:id="echoid-s726" xml:space="preserve">and BC, BE, BF will be
              <lb/>
            equal, and it has been proved that they are in a plane given in poſition, in
              <lb/>
            which plane is alſo the right line AD. </s>
            <s xml:id="echoid-s727" xml:space="preserve">The queſtion is then reduced to this,
              <lb/>
            Having two points E and F given, as alſo a right line AD in the ſame </s>
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