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-div37" type="section" level="1" n="37">
          <p>
            <s xml:id="echoid-s704" xml:space="preserve">
              <pb o="(20)" file="0032" n="32"/>
            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>
            <s xml:id="echoid-s710" xml:space="preserve"/>
          </p>
        </div>
        <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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