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-s149" xml:space="preserve">
              <emph style="sc">Case</emph>
            5th. </s>
            <s xml:id="echoid-s150" xml:space="preserve">Let one of the given circles include the other, and let be required
              <lb/>
            that the circle to be deſcribed be touched outwardly by one of the given circles,
              <lb/>
            and inwardly by the other.</s>
            <s xml:id="echoid-s151" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s152" xml:space="preserve">
              <emph style="sc">Limitation</emph>
            . </s>
            <s xml:id="echoid-s153" xml:space="preserve">Then it’s Diameter muſt not be given greater than the greater
              <lb/>
            ſegment of the right line, joining the centers of the given circles, which is in-
              <lb/>
            tercepted between the two concave circumferences of the ſaid circles, nor leſs
              <lb/>
            than the leſſer ſegment; </s>
            <s xml:id="echoid-s154" xml:space="preserve">viz. </s>
            <s xml:id="echoid-s155" xml:space="preserve">not greater than CD, nor leſs than MN.</s>
            <s xml:id="echoid-s156" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s157" xml:space="preserve">
              <emph style="sc">Case</emph>
            6th. </s>
            <s xml:id="echoid-s158" xml:space="preserve">Let the two given circles cut each other, and let it be required
              <lb/>
            that the circle to be deſcribed, and to be touched by them both, ſhall alſo be
              <lb/>
            included in each of them.</s>
            <s xml:id="echoid-s159" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s160" xml:space="preserve">
              <emph style="sc">Limitation</emph>
            . </s>
            <s xml:id="echoid-s161" xml:space="preserve">Then it’s Diameter muſt not be given greater than the ſeg-
              <lb/>
            ment of the right line, joining the centers of the given circles, intercepted by
              <lb/>
            their concave circumferences, which lies in the ſpace common to both the given
              <lb/>
            circles; </s>
            <s xml:id="echoid-s162" xml:space="preserve">viz. </s>
            <s xml:id="echoid-s163" xml:space="preserve">not greater than CD.</s>
            <s xml:id="echoid-s164" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s165" xml:space="preserve">
              <emph style="sc">There</emph>
            may be alſo three other Caſes of this Problem, when the given circles
              <lb/>
            cut each other; </s>
            <s xml:id="echoid-s166" xml:space="preserve">but becauſe they are ſimilar to the 1ſt, 2d, and 4th Caſes already
              <lb/>
            propoſed, and ſubject to juſt the ſame Limitations; </s>
            <s xml:id="echoid-s167" xml:space="preserve">except that which is ſimilar
              <lb/>
            to the 1ſt, which is ſubject to no Limitation at all, they are here omitted; </s>
            <s xml:id="echoid-s168" xml:space="preserve">as are
              <lb/>
            likewiſe thoſe Caſes where the given circles touch each other; </s>
            <s xml:id="echoid-s169" xml:space="preserve">becauſe they are
              <lb/>
            the ſame as the preceding, and ſolved in the ſame manner.</s>
            <s xml:id="echoid-s170" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div10" type="section" level="1" n="10">
          <head xml:id="echoid-head15" xml:space="preserve">
            <emph style="sc">The</emph>
            <emph style="sc">GENERAL</emph>
            <emph style="sc">Solution</emph>
          .</head>
          <p>
            <s xml:id="echoid-s171" xml:space="preserve">Join the given centers A and B, and where the Caſe requires, let AB be pro-
              <lb/>
            duced to meet the given circumferences in C and D: </s>
            <s xml:id="echoid-s172" xml:space="preserve">and let CI and DH be
              <lb/>
            taken equal to the given line Z: </s>
            <s xml:id="echoid-s173" xml:space="preserve">and let two circles be deſcribed; </s>
            <s xml:id="echoid-s174" xml:space="preserve">one with cen-
              <lb/>
            ter A and diſtance AI, and the other with center B and diſtance BH: </s>
            <s xml:id="echoid-s175" xml:space="preserve">and theſe
              <lb/>
            two circles will neceſſarily cut or touch each other by the Limitations given. </s>
            <s xml:id="echoid-s176" xml:space="preserve">Let
              <lb/>
            the point of concourſe be E: </s>
            <s xml:id="echoid-s177" xml:space="preserve">from E draw the right line EAF cutting the circle
              <lb/>
            whoſe center is A in F; </s>
            <s xml:id="echoid-s178" xml:space="preserve">as alſo EBG cutting the circle whoſe center is B in G:
              <lb/>
            </s>
            <s xml:id="echoid-s179" xml:space="preserve">then with center E and diſtance EF deſcribe a circle FK, this will be the circle
              <lb/>
            required: </s>
            <s xml:id="echoid-s180" xml:space="preserve">becauſe AF and AC are equal as alſo AI and AE; </s>
            <s xml:id="echoid-s181" xml:space="preserve">therefore FE
              <lb/>
            and CI are alſo equal: </s>
            <s xml:id="echoid-s182" xml:space="preserve">but CI was made equal to Z, therefore FE is equal to
              <lb/>
            Z. </s>
            <s xml:id="echoid-s183" xml:space="preserve">Again, becauſe BD and BG are equal, as alſo BH and BE, therefore DH
              <lb/>
            and EG are alſo equal: </s>
            <s xml:id="echoid-s184" xml:space="preserve">but DH was made equal to Z, therefore EG is equal to
              <lb/>
            Z. </s>
            <s xml:id="echoid-s185" xml:space="preserve">Hence it appears that the circle FK, paſſing through F will alſo paſs thro’
              <lb/>
            G, and likewiſe that it will alſo touch the given circles in F and G, becauſe EAF
              <lb/>
            and EBG are right lines paſſing through the centers.</s>
            <s xml:id="echoid-s186" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div11" type="section" level="1" n="11">
          <head xml:id="echoid-head16" xml:space="preserve">PROBLEM IV.</head>
          <p>
            <s xml:id="echoid-s187" xml:space="preserve">
              <emph style="sc">Having</emph>
            a given point A, and a given right line BC, it is required to draw a
              <lb/>
            circle, whoſe Radius ſhall be equal to a given line Z, which ſhall paſs through
              <lb/>
            the given point, and alſo touch the given line.</s>
            <s xml:id="echoid-s188" xml:space="preserve"/>
          </p>
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