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-div7" type="section" level="1" n="7">
          <head xml:id="echoid-head11" xml:space="preserve">PROBLEMS
            <lb/>
          CONCERNING
            <lb/>
          TANGENCIES.</head>
          <head xml:id="echoid-head12" xml:space="preserve">PROBLEM I.</head>
          <p>
            <s xml:id="echoid-s84" xml:space="preserve">THROUGH two given points A and B to deſcribe a circle whoſe radius
              <lb/>
            ſhall be equal to a given line Z.</s>
            <s xml:id="echoid-s85" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s86" xml:space="preserve">
              <emph style="sc">Limitation</emph>
            . </s>
            <s xml:id="echoid-s87" xml:space="preserve">2 Z muſt not be leſs than the diſtance of the points A and B.</s>
            <s xml:id="echoid-s88" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s89" xml:space="preserve">
              <emph style="sc">Construction</emph>
            . </s>
            <s xml:id="echoid-s90" xml:space="preserve">With the centers A and B, and diſtance Z, deſcribe two
              <lb/>
            arcs cutting or touching one another in the point E, ( which they will neceſſarily
              <lb/>
            do by the Limitation ) and E will be the center of the circle required.</s>
            <s xml:id="echoid-s91" xml:space="preserve"/>
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        <div xml:id="echoid-div8" type="section" level="1" n="8">
          <head xml:id="echoid-head13" xml:space="preserve">PROBLEM II.</head>
          <p>
            <s xml:id="echoid-s92" xml:space="preserve">
              <emph style="sc">Having</emph>
            two right lines AB CD given in poſition, it is required to draw 2
              <lb/>
            circle, whoſe Radius ſhall be equal to the given line Z, which ſhall alſo touch
              <lb/>
            both the given lines.</s>
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          </p>
          <p>
            <s xml:id="echoid-s94" xml:space="preserve">
              <emph style="sc">Case</emph>
            1ſt. </s>
            <s xml:id="echoid-s95" xml:space="preserve">Suppoſe AB and CD to be parallel.</s>
            <s xml:id="echoid-s96" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s97" xml:space="preserve">
              <emph style="sc">Limitation</emph>
            . </s>
            <s xml:id="echoid-s98" xml:space="preserve">2Z muſt be equal to the diſtance of the parallels, and the
              <lb/>
            conſtruction is evident.</s>
            <s xml:id="echoid-s99" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s100" xml:space="preserve">
              <emph style="sc">Case</emph>
            2d. </s>
            <s xml:id="echoid-s101" xml:space="preserve">Suppoſe AB and CD to be inclined to each other, let them be
              <lb/>
            produced till they meet in E, and let the angle BED be biſected by EH, and
              <lb/>
            through E draw EF perpendicular to ED, and equal to the given line Z; </s>
            <s xml:id="echoid-s102" xml:space="preserve">through
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            F draw FG parallel to EH, meeting ED in G, and through G draw GH paral-
              <lb/>
            lel to EF. </s>
            <s xml:id="echoid-s103" xml:space="preserve">I ſay that the circle deſcribed with H center, and HG radius,
              <lb/>
            touches the two given lines: </s>
            <s xml:id="echoid-s104" xml:space="preserve">it touches CD, becauſe EFGH is a </s>
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