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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          <head xml:id="echoid-head35" xml:space="preserve">SUPPLEMENT.</head>
          <head xml:id="echoid-head36" xml:space="preserve">PROBLEM I.</head>
          <p>
            <s xml:id="echoid-s596" xml:space="preserve">HAVING two points given A and B, to determine the Locus of the cen-
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            ters of all ſuch circles as may be drawn through A and B.</s>
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              <emph style="sc">Join</emph>
            AB and biſect it in the point C, and through C, draw a perpendicular
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            to it CE, and continue it both ways in infinitum, and it is evident that this line
              <lb/>
            CE will be the Locus required.</s>
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          <head xml:id="echoid-head37" xml:space="preserve">PROBLEM II.</head>
          <p>
            <s xml:id="echoid-s600" xml:space="preserve">
              <emph style="sc">Having</emph>
            two right lines given AB and CD, to determine the Locus of the
              <lb/>
            centers of all ſuch circles as may be drawn touching both the ſaid lines.</s>
            <s xml:id="echoid-s601" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s602" xml:space="preserve">
              <emph style="sc">Case</emph>
            1ft. </s>
            <s xml:id="echoid-s603" xml:space="preserve">Suppoſe AB and CD to be parallel; </s>
            <s xml:id="echoid-s604" xml:space="preserve">then drawing GI perpendicu-
              <lb/>
            lar to them both; </s>
            <s xml:id="echoid-s605" xml:space="preserve">biſect it in H, and through H draw EHF parallel to the two
              <lb/>
            given lines, and it will be the Locus required.</s>
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            <s xml:id="echoid-s607" xml:space="preserve">
              <emph style="sc">Case</emph>
            2d. </s>
            <s xml:id="echoid-s608" xml:space="preserve">Suppoſe the given lines being produced meet each other in E,
              <lb/>
            then biſecting the angle BED by the line EHF, this line EHF will be the Locus
              <lb/>
            required.</s>
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          <head xml:id="echoid-head38" xml:space="preserve">PROBLEM III.</head>
          <p>
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              <emph style="sc">Having</emph>
            two circles given whoſe centers are A and B; </s>
            <s xml:id="echoid-s611" xml:space="preserve">to determine the Locus
              <lb/>
            of the centers of all ſuch circles as ſhall touch the two given ones.</s>
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              <emph style="sc">Cases</emph>
            1ſt and 2d. </s>
            <s xml:id="echoid-s614" xml:space="preserve">Suppoſe it be required that the circles be touched outwardly
              <lb/>
            by both the given ones; </s>
            <s xml:id="echoid-s615" xml:space="preserve">or that they be touched inwardly by both the given
              <lb/>
            ones.</s>
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              <emph style="sc">Then</emph>
            joining the centers A and B, let AB cut the circumferences in C and
              <lb/>
            D and produced in P and O: </s>
            <s xml:id="echoid-s618" xml:space="preserve">let CD which is intercepted between the convex
              <lb/>
            circumſerences be biſected in E: </s>
            <s xml:id="echoid-s619" xml:space="preserve">ſet off from E towards B the center of </s>
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