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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[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.
[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.
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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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            <s xml:id="echoid-s598" xml:space="preserve">
              <emph style="sc">Join</emph>
            AB and biſect it in the point C, and through C, draw a perpendicular
              <lb/>
            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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        <div xml:id="echoid-div31" type="section" level="1" n="31">
          <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>
            <s xml:id="echoid-s606" xml:space="preserve"/>
          </p>
          <p>
            <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>
            <s xml:id="echoid-s610" xml:space="preserve">
              <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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            <s xml:id="echoid-s617" xml:space="preserve">
              <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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