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-div32" type="section" level="1" n="32">
          <p>
            <s xml:id="echoid-s619" xml:space="preserve">
              <pb o="(15)" file="0027" n="27"/>
            greater of the given circles the line EH = the difference of the Radii of the given
              <lb/>
            circles, and with A and B as Foci and EH Tranverſe Axis, let two oppoſite
              <lb/>
            Hyperbolas be deſcribed KEK and LHL: </s>
            <s xml:id="echoid-s620" xml:space="preserve">then I ſay that KEK will be the
              <lb/>
            locus of the centers of all the circles which can be drawn ſo as to be touched
              <lb/>
            outwardly by both the given circles; </s>
            <s xml:id="echoid-s621" xml:space="preserve">and LHL will be the locus of the centers
              <lb/>
            of all the circles which can be drawn ſo as to be touched inwardly by both the
              <lb/>
            given circles. </s>
            <s xml:id="echoid-s622" xml:space="preserve">For taking any point K in the Hyperbola KEK, and drawing KA
              <lb/>
            and KB, let theſe lines cut the given circumferences in F and G reſpectively:
              <lb/>
            </s>
            <s xml:id="echoid-s623" xml:space="preserve">and make KQ = KA: </s>
            <s xml:id="echoid-s624" xml:space="preserve">then from the nature of the curve QB = EH, but by con-
              <lb/>
            ſtruction EH = BG - AF, therefore QB = BG - AF and hence QG = AF, and
              <lb/>
            then taking equals from equals KG = KF, which is a demonſtration of the 1ſt
              <lb/>
            Caſe.</s>
            <s xml:id="echoid-s625" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s626" xml:space="preserve">
              <emph style="sc">Then</emph>
            with regard to the 2d Caſe, taking any point L in the Hyperbola
              <lb/>
            LHL, and drawing LB and LA and producing them to meet the concave cir-
              <lb/>
            cumferences in M and N, let alſo LR be taken equal to LB; </s>
            <s xml:id="echoid-s627" xml:space="preserve">then from the pro-
              <lb/>
            perty of the curve AR = EH, but EH (by conſtruction) = BM - NA; </s>
            <s xml:id="echoid-s628" xml:space="preserve">there-
              <lb/>
            fore AR = BM - NA, and NR = BM, and then adding equals to equals LN =
              <lb/>
            LM, which is a demonſtration of the 2d Caſe.</s>
            <s xml:id="echoid-s629" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s630" xml:space="preserve">
              <emph style="sc">Case</emph>
            3d. </s>
            <s xml:id="echoid-s631" xml:space="preserve">Suppoſe it be required that the circles to be deſcribed be touched
              <lb/>
            outwardly by one of the given circles, and inwardly by the other.</s>
            <s xml:id="echoid-s632" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s633" xml:space="preserve">Then drawing AB, let it cut the convex circumferences in C and D, and the
              <lb/>
            concave ones in P and O, biſect PD in E, and from E towards B ſet off EH =
              <lb/>
            the ſum of the given radii. </s>
            <s xml:id="echoid-s634" xml:space="preserve">Then with A and B foci and EH tranſverſe axis,
              <lb/>
            let two oppoſite hyperbolas be deſcribed KEK and LHL: </s>
            <s xml:id="echoid-s635" xml:space="preserve">and KEK will be the
              <lb/>
            locus of the centers of the circles which are touched inwardly by the circle A
              <lb/>
            and outwardly by the circle B; </s>
            <s xml:id="echoid-s636" xml:space="preserve">and LHL will be the locus of the centers of
              <lb/>
            thoſe circles which are touched inwardly by B and outwardly by A. </s>
            <s xml:id="echoid-s637" xml:space="preserve">The demon-
              <lb/>
            ſtration mutatis mutandis is the ſame as before.</s>
            <s xml:id="echoid-s638" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s639" xml:space="preserve">
              <emph style="sc">Case</emph>
            4th. </s>
            <s xml:id="echoid-s640" xml:space="preserve">Suppoſe the given circle A to include B, and it be required that
              <lb/>
            the circles to be deſcribed be touched outwardly by them both.</s>
            <s xml:id="echoid-s641" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s642" xml:space="preserve">Let AB cut the circumferences in C and D, P and O: </s>
            <s xml:id="echoid-s643" xml:space="preserve">and biſecting CD in
              <lb/>
            I, and ſetting off from I towards P, IL = the ſum of the ſemidiameters of the
              <lb/>
            given circles, and with A and B foci, and IL tranſverſe axis, deſcribing an
              <lb/>
            ellipſe LKI, it will be the locus of the centers of the circles required. </s>
            <s xml:id="echoid-s644" xml:space="preserve">For
              <lb/>
            taking any point K in the ellipſe, and drawing AK and BK, let AK be con-
              <lb/>
            tinued to meet one of the given circumferences in G, and let BK meet the other
              <lb/>
            F. </s>
            <s xml:id="echoid-s645" xml:space="preserve">Then from the property of the curve AK + BK = IL = AG + BF (by </s>
          </p>
        </div>
      </text>
    </echo>
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