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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          <p>
            <s xml:id="echoid-s798" xml:space="preserve">
              <pb o="(24)" file="0036" n="36"/>
            Thercfore there are given three points H, M, D, as likewiſe a plane AB,
              <lb/>
            or AC, through which points the ſphere is to paſs, and alſo touch the given
              <lb/>
            plane. </s>
            <s xml:id="echoid-s799" xml:space="preserve">Hence it appears that this Problem is reduced to the IId of this
              <lb/>
            Supplement.</s>
            <s xml:id="echoid-s800" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s801" xml:space="preserve">
              <emph style="sc">Before</emph>
            we proceed, the following eaſy Lemmas muſt be premiſed.</s>
            <s xml:id="echoid-s802" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div44" type="section" level="1" n="44">
          <head xml:id="echoid-head51" xml:space="preserve">LEMMA I.</head>
          <p>
            <s xml:id="echoid-s803" xml:space="preserve">
              <emph style="sc">Let</emph>
            there be a circle BCD, and a point E taken without it, and iſ from
              <lb/>
            E a line EDOB be drawn to paſs through the center, and another line ECA
              <lb/>
            to cut it any ways; </s>
            <s xml:id="echoid-s804" xml:space="preserve">we know from the Elements that the rectangle AEC is
              <lb/>
            equal to the rectangle BED. </s>
            <s xml:id="echoid-s805" xml:space="preserve">Let us now ſuppoſe a ſphere whoſe center is O,
              <lb/>
            and one of whoſe great circles is ACDB; </s>
            <s xml:id="echoid-s806" xml:space="preserve">if from the ſame point E a line
              <lb/>
            ECA be any-how drawn to meet the ſpherical ſurface in the points C and A,
              <lb/>
            I ſay the rectangle AEC will ſtill be equal to the rectangle BED. </s>
            <s xml:id="echoid-s807" xml:space="preserve">For if we
              <lb/>
            ſuppoſe the circle and right line ECA to revolve upon EDB as an immove-
              <lb/>
            able axis, the lines EC and EA will not be changed, becauſe the points C
              <lb/>
            and A deſcribe circles whoſe planes are perpendicular to that axis; </s>
            <s xml:id="echoid-s808" xml:space="preserve">and
              <lb/>
            therefore the rectangle AEC will in any plane be ſtill equal to the rectangle
              <lb/>
            BED.</s>
            <s xml:id="echoid-s809" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div45" type="section" level="1" n="45">
          <head xml:id="echoid-head52" xml:space="preserve">LEMMA II.</head>
          <p>
            <s xml:id="echoid-s810" xml:space="preserve">
              <emph style="sc">By</emph>
            the ſame method of reaſoning, the Vth Lemma immediately preceed-
              <lb/>
            ing Problem XIII, in the Treatiſe of Circular Tangencies, may be extended
              <lb/>
            alſo to ſpheres, viz. </s>
            <s xml:id="echoid-s811" xml:space="preserve">that in any plane (ſee the Figures belonging to that
              <lb/>
            Lemma) MG X MB = MH X MA. </s>
            <s xml:id="echoid-s812" xml:space="preserve">And alſo that MF X MC = ME X MI.</s>
            <s xml:id="echoid-s813" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div46" type="section" level="1" n="46">
          <head xml:id="echoid-head53" xml:space="preserve">LEMMA III.</head>
          <p>
            <s xml:id="echoid-s814" xml:space="preserve">
              <emph style="sc">Let</emph>
            there be two ſpheres YN, XM, through whoſe centers let the right
              <lb/>
            line RYNXMU paſs, and let it be as the radius YN to the radius XM, ſo
              <lb/>
            YU to XU; </s>
            <s xml:id="echoid-s815" xml:space="preserve">and from the point U let a line UTS be drawn in any plane,
              <lb/>
            and let the rectangle S U T be equal to the rectangle RUM; </s>
            <s xml:id="echoid-s816" xml:space="preserve">I fay that if
              <lb/>
            any ſphere OTS be deſcribed to paſs through the points T and S, and to
              <lb/>
            touch one of the given ſpheres XM as in O, it will alſo touch the other
              <lb/>
            given ſphere YN. </s>
            <s xml:id="echoid-s817" xml:space="preserve">For joining UO, and producing it to meet the ſurface of
              <lb/>
            the ſphere OTS in Q; </s>
            <s xml:id="echoid-s818" xml:space="preserve">the rectangle QUO = the rectangle SUT, by
              <lb/>
            Lemma I. </s>
            <s xml:id="echoid-s819" xml:space="preserve">but the rectangle SUT = the rectangle RUM. </s>
            <s xml:id="echoid-s820" xml:space="preserve">by </s>
          </p>
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