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

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        <div xml:id="echoid-div34" type="section" level="1" n="34">
          <pb o="(17)" file="0029" n="29"/>
          <p>
            <s xml:id="echoid-s673" xml:space="preserve">
              <emph style="sc">Case</emph>
            3d. </s>
            <s xml:id="echoid-s674" xml:space="preserve">Suppoſe the given point A to lie in the given circle, whoſe center
              <lb/>
            is B.</s>
            <s xml:id="echoid-s675" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s676" xml:space="preserve">
              <emph style="sc">Then</emph>
            joining AB and continuing it to meet the given circumference in C and
              <lb/>
            O, biſect AC in E, and from E towards O ſetting off EH = BC the given
              <lb/>
            Radius, and with A and B Foci and EH tranſverſe Axis deſcribing an Ellipſe
              <lb/>
            EKH, it will evidently be the Locus required.</s>
            <s xml:id="echoid-s677" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div35" type="section" level="1" n="35">
          <head xml:id="echoid-head41" xml:space="preserve">PROBLEM VI.</head>
          <p>
            <s xml:id="echoid-s678" xml:space="preserve">
              <emph style="sc">Having</emph>
            a right line BC given, and alſo a circle whoſe center is A, to deter-
              <lb/>
            mine the Locus of the centers of the circles which ſhall be touched both by the
              <lb/>
            given right line and alſo by the given circle.</s>
            <s xml:id="echoid-s679" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s680" xml:space="preserve">
              <emph style="sc">There</emph>
            are three Caſes, but they are all comprchended under one general
              <lb/>
            ſolution.</s>
            <s xml:id="echoid-s681" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s682" xml:space="preserve">
              <emph style="sc">Case</emph>
            1ſt. </s>
            <s xml:id="echoid-s683" xml:space="preserve">Let the given right line be without the given circle, and let it be
              <lb/>
            required that the circles to be deſcribed be touched outwardly by the given circle.</s>
            <s xml:id="echoid-s684" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s685" xml:space="preserve">
              <emph style="sc">Case</emph>
            2d. </s>
            <s xml:id="echoid-s686" xml:space="preserve">Let the given right line be without the given circle, and let it be
              <lb/>
            required that the circles to be deſcribed, be touched inwardly by the given
              <lb/>
            circle.</s>
            <s xml:id="echoid-s687" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s688" xml:space="preserve">
              <emph style="sc">Case</emph>
            3d. </s>
            <s xml:id="echoid-s689" xml:space="preserve">Let the given right line be within the given circle, and then the
              <lb/>
            circles to be deſcribed muſt be touched outwardly by the given circle.</s>
            <s xml:id="echoid-s690" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div36" type="section" level="1" n="36">
          <head xml:id="echoid-head42" xml:space="preserve">
            <emph style="sc">General</emph>
            <emph style="sc">Solution.</emph>
          </head>
          <p>
            <s xml:id="echoid-s691" xml:space="preserve">
              <emph style="sc">From</emph>
            the given center A let fall a perpendicular AG to the given line BC,
              <lb/>
            which meets the given circumference in D [or in Caſes 2d and 3d is produced
              <lb/>
            to meet it in D] and biſecting DG in F, and ſetting off FM = FA (which is the
              <lb/>
            ſame thing as making GM = AD the given Radius) and through M drawing
              <lb/>
            MLK parallel to the given line BC, with A Focus and LK Directrix deſcribe a
              <lb/>
            Parabola, and it will be the Locus of the centers of the circles required; </s>
            <s xml:id="echoid-s692" xml:space="preserve">for
              <lb/>
            from the property of the Curve FA = FM, and adding equals to equals, or
              <lb/>
            ſubtracting equals from equals, as the Caſe requires, FD = FG.</s>
            <s xml:id="echoid-s693" xml:space="preserve"/>
          </p>
        </div>
      </text>
    </echo>
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