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

Table of contents

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[21.] LEMMA II.
Page: 20 ((8))
[22.] LEMMA III.
Page: 20 ((8))
[23.] PROBLEM X.
Page: 20 ((8))
[24.] PROBLEM XI.
Page: 21 ((9))
[25.] PROBLEM XII .
Page: 22 ((10))
[26.] LEMMA IV.
Page: 23 ((11))
[27.] LEMMA V.
Page: 23 ((11))
[28.] PROBLEM XIII.
Page: 23 ((11))
[29.] PROBLEM XIV.
Page: 24 ((12))
[30.] SUPPLEMENT. PROBLEM I.
Page: 26 ((14))
[31.] PROBLEM II.
Page: 26 ((14))
[32.] PROBLEM III.
Page: 26 ((14))
[33.] PROBLEM IV.
Page: 28 ((16))
[34.] PROBLEM V.
Page: 28 ((16))
[35.] PROBLEM VI.
Page: 29 ((17))
[36.] General Solution.
Page: 29 ((17))
[37.] A SECOND SUPPLEMENT, BEING Monſ. DE FERMAT’S Treatiſe on Spherical Tangencies. PROBLEM I.
Page: 31 ((19))
[38.] PROBLEM II.
Page: 32 ((20))
[39.] PROBLEM III.
Page: 33 ((21))
[40.] PROBLEM IV.
Page: 33 ((21))
[41.] PROBLEM V.
Page: 34 ((22))
[42.] PROBLEM VI.
Page: 34 ((22))
[43.] PROBLEM VII.
Page: 35 ((23))
[44.] LEMMA I.
Page: 36 ((24))
[45.] LEMMA II.
Page: 36 ((24))
[46.] LEMMA III.
Page: 36 ((24))
[47.] LEMMA IV.
Page: 37 ((25))
[48.] LEMMA V.
Page: 37 ((25))
[49.] PROBLEM VIII.
Page: 38 ((26))
[50.] PROBLEM IX.
Page: 39 ((27))
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            <s xml:id="echoid-s538" xml:space="preserve">
              <pb o="(12)" file="0024" n="24"/>
            MA x MH: </s>
            <s xml:id="echoid-s539" xml:space="preserve">but MA x MH = MD x MN by Conſtruction; </s>
            <s xml:id="echoid-s540" xml:space="preserve">therefore MB x
              <lb/>
            MG = MD x MN, and the points B, G, N, D, are in a circle. </s>
            <s xml:id="echoid-s541" xml:space="preserve">But the point
              <lb/>
            G is alſo in the circle L; </s>
            <s xml:id="echoid-s542" xml:space="preserve">therefore theſe circles either touch or cut each other in
              <lb/>
            the point G. </s>
            <s xml:id="echoid-s543" xml:space="preserve">Now the circles BND and BCI touch one another in B by con-
              <lb/>
            ſtruction; </s>
            <s xml:id="echoid-s544" xml:space="preserve">therefore the ſegment BC is ſimilar to the ſegment BG; </s>
            <s xml:id="echoid-s545" xml:space="preserve">and alſo by
              <lb/>
            conſtruction the ſegment BC is ſimilar to the ſegment FG; </s>
            <s xml:id="echoid-s546" xml:space="preserve">and therefore the
              <lb/>
            ſegment FG is ſimilar to the ſegment BG; </s>
            <s xml:id="echoid-s547" xml:space="preserve">and hence the circles FGE and BGD
              <lb/>
            touch one another in the point G.</s>
            <s xml:id="echoid-s548" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s549" xml:space="preserve">
              <emph style="sc">The</emph>
            Caſes are three.</s>
            <s xml:id="echoid-s550" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s551" xml:space="preserve">
              <emph style="sc">Case</emph>
            Iſt. </s>
            <s xml:id="echoid-s552" xml:space="preserve">Iſ the circle be required to touch and include both the given ones;
              <lb/>
            </s>
            <s xml:id="echoid-s553" xml:space="preserve">then M muſt be taken in KL produced; </s>
            <s xml:id="echoid-s554" xml:space="preserve">and MN muſt be taken a fourth pro-
              <lb/>
            portional to MD, MA, MH, A being the moſt diſtant point of interſection in
              <lb/>
            the circle K, and H the neareſt point of interſection in the circle L.</s>
            <s xml:id="echoid-s555" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s556" xml:space="preserve">
              <emph style="sc">Case</emph>
            2d. </s>
            <s xml:id="echoid-s557" xml:space="preserve">If the circle be required to touch both the given ones externally;
              <lb/>
            </s>
            <s xml:id="echoid-s558" xml:space="preserve">then alſo M muſt be taken in KL produced; </s>
            <s xml:id="echoid-s559" xml:space="preserve">and MN taken a fourth proportional
              <lb/>
            to MD, MA, MH, A being the neareſt point of interſection in the circle K,
              <lb/>
            and H the moſt diſtant in the circle L.</s>
            <s xml:id="echoid-s560" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s561" xml:space="preserve">
              <emph style="sc">Case</emph>
            3d. </s>
            <s xml:id="echoid-s562" xml:space="preserve">If the circle be required to touch and include the circle K, and to
              <lb/>
            touch L externally; </s>
            <s xml:id="echoid-s563" xml:space="preserve">then M muſt be taken in KL itſelf; </s>
            <s xml:id="echoid-s564" xml:space="preserve">and MN a fourth pro-
              <lb/>
            portional to MD, MA, MH, A being the moſt diſtant point in K, and H the
              <lb/>
            neareſt in L.</s>
            <s xml:id="echoid-s565" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div29" type="section" level="1" n="29">
          <head xml:id="echoid-head34" xml:space="preserve">PROBLEM XIV.</head>
          <p>
            <s xml:id="echoid-s566" xml:space="preserve">
              <emph style="sc">Having</emph>
            three circles given whoſe centers are A, B, and D; </s>
            <s xml:id="echoid-s567" xml:space="preserve">to draw a fourth
              <lb/>
            which ſhall touch all three.</s>
            <s xml:id="echoid-s568" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s569" xml:space="preserve">
              <emph style="sc">Let</emph>
            that whoſe center is A be called the 1ſt, that whoſe center is B the 2d,
              <lb/>
            and that whoſe center is D the 3d. </s>
            <s xml:id="echoid-s570" xml:space="preserve">Then with center B, and radius equal to
              <lb/>
            the differcnce, or ſum, as the caſe requires, of the ſemidiameters of the 1ſt and
              <lb/>
            2d circles, let an auxiliary circle be deſcribed; </s>
            <s xml:id="echoid-s571" xml:space="preserve">and likewiſe with D center, and
              <lb/>
            radius equal to the difference, or ſum, as the caſe requires, of the ſemidiameters
              <lb/>
            of the 1ſt and 3d circles, let another auxiliary circle be deſcribed; </s>
            <s xml:id="echoid-s572" xml:space="preserve">and laſtly by
              <lb/>
            the preceding Problem draw a circle which ſhall touch the two auxiliary ones,
              <lb/>
            and likewiſe paſs through the point A which is the center of the firſt circle.
              <lb/>
            </s>
            <s xml:id="echoid-s573" xml:space="preserve">Let the center of this laſt deſcribed circle be E, and the ſame point E will like-
              <lb/>
            wiſe be the center of the circle required; </s>
            <s xml:id="echoid-s574" xml:space="preserve">as will appear by adding equals to
              <lb/>
            equals, or taking equals from equals, as the caſe requires.</s>
            <s xml:id="echoid-s575" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s576" xml:space="preserve">
              <emph style="sc">The</emph>
            Caſes are theſe.</s>
            <s xml:id="echoid-s577" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s578" xml:space="preserve">
              <emph style="sc">Case</emph>
            1ſt. </s>
            <s xml:id="echoid-s579" xml:space="preserve">Iſ it be required that the circle ſhould touch and include all the other
              <lb/>
            three; </s>
            <s xml:id="echoid-s580" xml:space="preserve">then let A be the center of the greateſt given circle, B of the next, and D
              <lb/>
            of the leaſt: </s>
            <s xml:id="echoid-s581" xml:space="preserve">and let BG = the difference of the ſemidiameters of the 1ſt and </s>
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