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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            ſtruction:) </s>
            <s xml:id="echoid-s646" xml:space="preserve">hence by ſubſtraction BK = KG + BF, and by ſubtraction again
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            FK = KG.</s>
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          <p>
            <s xml:id="echoid-s648" xml:space="preserve">
              <emph style="sc">Case</emph>
            5th. </s>
            <s xml:id="echoid-s649" 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 A and inwardly by B.</s>
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          <p>
            <s xml:id="echoid-s651" xml:space="preserve">
              <emph style="sc">Then</emph>
            let AB cut the circumferences in C and D, P and O: </s>
            <s xml:id="echoid-s652" xml:space="preserve">and biſecting
              <lb/>
            CO in I, and ſetting off from I towards P, IL = the difference of the ſemidia-
              <lb/>
            meters of the given circles, and with A and B foci and IL tranſverſe axis de-
              <lb/>
            ſcribing an ellipſe LKI, it will be the locus of the centers of the circles deſcribed,
              <lb/>
            and the demonſtration, mutatis mutandis, is the ſame as in the laſt caſe.</s>
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          <p>
            <s xml:id="echoid-s654" xml:space="preserve">
              <emph style="sc">Cases</emph>
            6th and 7th. </s>
            <s xml:id="echoid-s655" xml:space="preserve">Suppoſe the two given circles cut each other, and it be
              <lb/>
            required that the circles to be deſcribed either be touched and included in them
              <lb/>
            both, or be touched by them both and at the ſame time include them both.</s>
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          </p>
          <p>
            <s xml:id="echoid-s657" xml:space="preserve">
              <emph style="sc">These</emph>
            two caſes are ſimilar to caſes 1ſt and 2d, and as there, ſo alſo here,
              <lb/>
            the tranſverſe axis of the two oppoſite hyperbolas, which are the loci required,
              <lb/>
            muſt be taken = the difference of the ſemidiameters of the given circles. </s>
            <s xml:id="echoid-s658" xml:space="preserve">The
              <lb/>
            demonſtration is ſo alike, it need not be repeated.</s>
            <s xml:id="echoid-s659" xml:space="preserve"/>
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        <div xml:id="echoid-div33" type="section" level="1" n="33">
          <head xml:id="echoid-head39" xml:space="preserve">PROBLEM IV.</head>
          <p>
            <s xml:id="echoid-s660" xml:space="preserve">
              <emph style="sc">Having</emph>
            a given point A, and a given right line BC, to determine the locus
              <lb/>
            of the centers of thoſe circles which ſhall paſs through A and touch BC.</s>
            <s xml:id="echoid-s661" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s662" xml:space="preserve">
              <emph style="sc">From</emph>
            A draw AG perpendicular to BC, then with focus A and directrix BC
              <lb/>
            let a parabola be deſcribed, and it will be the locus required; </s>
            <s xml:id="echoid-s663" xml:space="preserve">for by the propert
              <lb/>
            of the curve FA always equals FG drawn perpendicular to the directrix.</s>
            <s xml:id="echoid-s664" xml:space="preserve"/>
          </p>
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          <head xml:id="echoid-head40" xml:space="preserve">PROBLEM V.</head>
          <p>
            <s xml:id="echoid-s665" xml:space="preserve">
              <emph style="sc">Having</emph>
            a given point A, and a given circle whoſe center is B, to determine
              <lb/>
            the locus of the centers of all thoſe circles, which paſs through A, and at the
              <lb/>
            ſame time are touched by the given circle.</s>
            <s xml:id="echoid-s666" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s667" xml:space="preserve">
              <emph style="sc">Cases</emph>
            1ſt and 2d. </s>
            <s xml:id="echoid-s668" xml:space="preserve">Suppoſe the point A to lie out of the given circle, and
              <lb/>
            it be required that the circles to be deſcribed be either touched outwardly by the
              <lb/>
            given circle, or inwardly by it.</s>
            <s xml:id="echoid-s669" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s670" xml:space="preserve">
              <emph style="sc">Let</emph>
            AB be drawn, and let it cut the given circumference where it is convex
              <lb/>
            towards A in the point C, and where it is concave in the point O: </s>
            <s xml:id="echoid-s671" xml:space="preserve">then biſecting
              <lb/>
            AC in E, and ſetting off from E towards B, EH = BC the given radius, and
              <lb/>
            with A and B foci and EH tranſverſe Axis deſcribing two oppoſite Hyperbolas
              <lb/>
            KEK and LHL, it is evident that KEK will be the locus of the centers of thoſe
              <lb/>
            circles which paſs thrugh A and are touched outwardly by the given circle, and
              <lb/>
            LHL will be the locus of the centers of thoſe circles which paſs through A and
              <lb/>
            are touched inwardly by the given circle.</s>
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