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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[51.] PROBLEM X.
[52.] PROBLEM XI.
[53.] PROBLEM XII.
[54.] PROBLEM XIII.
[55.] PROBLEM XIV.
[56.] PROBLEM XV.
[57.] Synopſis of the PROBLEMS.
[58.] THE TWO BOOKS OF APOLLONIUS PERGÆUS, CONCERNING DETERMINATE SECTION, As they have been Reſtored by WILLEBRORDUS SNELLIUS. By JOHN LAWSON, B. D. Rector of Swanſcombe, Kent. TO WHICH ARE ADDED, THE SAME TWO BOOKS, BY WILLIAM WALES, BEING AN ENTIRE NEW WORK. LONDON: Printed by G. BIGG, Succeſſor to D. LEACH. And ſold by B. White, in Fleet-Street; L. Davis, in Holborne; J. Nourse, in the Strand; and T. Payne, near the Mews-Gate. MDCC LXXII.
[59.] ADVERTISEMENT.
[60.] EXTRACT from PAPPUS's Preface to his Seventh Book in Dr. HALLEY's Tranſlation. DE SECTIONE DETERMINATA II.
[61.] THE PREFACE.
[62.] PROBLEMS CONCERNING DETERMINATE SECTION. PROBLEM I.
[63.] LEMMA I.
[64.] LEMMA II.
[65.] LEMMA III.
[66.] PROBLEM II.
[67.] LEMMA IV.
[68.] LEMMA V.
[69.] PROBLEM III.
[70.] PROBLEM IV.
[71.] DETERMINATE SECTION. BOOK I. PROBLEM I. (Fig. 1.)
[72.] PROBLEM II. (Fig. 2 and 3.)
[73.] PROBLEM III. (Fig. 4. and 5.)
[74.] PROBLEM IV. (Fig. 6. 7. and 8.)
[75.] PROBLEM V. (Fig. 9. 10. 11. 12. 13. 14. 15. 16.)
[76.] PROBLEM VI. (Fig. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.)
[77.] THE END OF BOOK I.
[78.] DETERMINATE SECTION. BOOK II. LEMMA I.
[79.] LEMMA II.
[80.] LEMMA III.
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        <div xml:id="echoid-div26" type="section" level="1" n="26">
          <head xml:id="echoid-head31" xml:space="preserve">LEMMA IV.</head>
          <p>
            <s xml:id="echoid-s477" xml:space="preserve">
              <emph style="sc">Having</emph>
            two circles ABCI and EFGH given, it is required to find a point
              <lb/>
            M, in the line joining their centers, or in that line continued, ſuch, that any
              <lb/>
            line drawn through the ſaid point M, cutting both the circles, ſhall always cut
              <lb/>
            off ſimilar ſegments.</s>
            <s xml:id="echoid-s478" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s479" xml:space="preserve">
              <emph style="sc">Let</emph>
            the line KL joining the centers be ſo cut or produced, that KM may be
              <lb/>
            to LM in the given ratio of the Radii AK to EL [in the caſe of producing
              <lb/>
            KL we muſt make it as AK-EL: </s>
            <s xml:id="echoid-s480" xml:space="preserve">EL:</s>
            <s xml:id="echoid-s481" xml:space="preserve">: KL: </s>
            <s xml:id="echoid-s482" xml:space="preserve">LM, for then by compoſition it
              <lb/>
            will be AK: </s>
            <s xml:id="echoid-s483" xml:space="preserve">EL:</s>
            <s xml:id="echoid-s484" xml:space="preserve">: KM: </s>
            <s xml:id="echoid-s485" xml:space="preserve">LM] and then I ſay that the point M will be the
              <lb/>
            point required. </s>
            <s xml:id="echoid-s486" xml:space="preserve">For from it drawing any line MGFCB cutting the circle
              <lb/>
            ABCI in B and C, and the circle EFGH in F and G; </s>
            <s xml:id="echoid-s487" xml:space="preserve">and let B and F be the
              <lb/>
            correſpondent points moſt diſtant from M, and C and G the correſpondent
              <lb/>
            points that are nearer; </s>
            <s xml:id="echoid-s488" xml:space="preserve">and let be joined BK, CK, FL, GL, and thereby two
              <lb/>
            triangles will be formed BKC, FLG. </s>
            <s xml:id="echoid-s489" xml:space="preserve">Now becauſe by Conſtri
              <unsure/>
            ction KM: </s>
            <s xml:id="echoid-s490" xml:space="preserve">LM
              <lb/>
            :</s>
            <s xml:id="echoid-s491" xml:space="preserve">: KB: </s>
            <s xml:id="echoid-s492" xml:space="preserve">LF, KB and LF will be parallel by Euc. </s>
            <s xml:id="echoid-s493" xml:space="preserve">VI. </s>
            <s xml:id="echoid-s494" xml:space="preserve">7. </s>
            <s xml:id="echoid-s495" xml:space="preserve">and for the ſame rea-
              <lb/>
            ſon KC and LG will be parallel; </s>
            <s xml:id="echoid-s496" xml:space="preserve">and therefore the triangles BKC and FLG
              <lb/>
            will be equiangled, and hence the ſegment BC will be ſimilar to the ſegment FG.</s>
            <s xml:id="echoid-s497" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div27" type="section" level="1" n="27">
          <head xml:id="echoid-head32" xml:space="preserve">LEMMA V.</head>
          <p>
            <s xml:id="echoid-s498" xml:space="preserve">
              <emph style="sc">The</emph>
            point M being found as in the preceding Lemma, I ſay that it is a pro-
              <lb/>
            perty of the ſaid point, that MG x MB = MH x MA: </s>
            <s xml:id="echoid-s499" xml:space="preserve">as alſo that MF x
              <lb/>
            MC = ME x MI.</s>
            <s xml:id="echoid-s500" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s501" xml:space="preserve">
              <emph style="sc">For</emph>
            joining CI and GH, it is evident that theſe lines will alſo be parallel.
              <lb/>
            </s>
            <s xml:id="echoid-s502" xml:space="preserve">Hence MI: </s>
            <s xml:id="echoid-s503" xml:space="preserve">MC:</s>
            <s xml:id="echoid-s504" xml:space="preserve">: MH: </s>
            <s xml:id="echoid-s505" xml:space="preserve">MG, but MI: </s>
            <s xml:id="echoid-s506" xml:space="preserve">MC:</s>
            <s xml:id="echoid-s507" xml:space="preserve">: MB: </s>
            <s xml:id="echoid-s508" xml:space="preserve">MA, therefore
              <lb/>
            MH: </s>
            <s xml:id="echoid-s509" xml:space="preserve">MG:</s>
            <s xml:id="echoid-s510" xml:space="preserve">: MB: </s>
            <s xml:id="echoid-s511" xml:space="preserve">MA, and MG x MB = MH x MA. </s>
            <s xml:id="echoid-s512" xml:space="preserve">Again MH: </s>
            <s xml:id="echoid-s513" xml:space="preserve">MG
              <lb/>
            :</s>
            <s xml:id="echoid-s514" xml:space="preserve">: MF: </s>
            <s xml:id="echoid-s515" xml:space="preserve">ME, but MH: </s>
            <s xml:id="echoid-s516" xml:space="preserve">MG:</s>
            <s xml:id="echoid-s517" xml:space="preserve">: MI: </s>
            <s xml:id="echoid-s518" xml:space="preserve">MC, therefore MF: </s>
            <s xml:id="echoid-s519" xml:space="preserve">ME:</s>
            <s xml:id="echoid-s520" xml:space="preserve">: MI: </s>
            <s xml:id="echoid-s521" xml:space="preserve">MC
              <lb/>
            and MF x MC = ME x MI.</s>
            <s xml:id="echoid-s522" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div28" type="section" level="1" n="28">
          <head xml:id="echoid-head33" xml:space="preserve">PROBLEM XIII.</head>
          <p>
            <s xml:id="echoid-s523" xml:space="preserve">
              <emph style="sc">Having</emph>
            two circles given in magnitude and poſition, whoſe centers are K
              <lb/>
            and L, and alſo a point D; </s>
            <s xml:id="echoid-s524" xml:space="preserve">to draw a circle which ſhall touch the two given
              <lb/>
            ones, and alſo paſs through the point D.</s>
            <s xml:id="echoid-s525" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s526" xml:space="preserve">
              <emph style="sc">Join</emph>
            the given centers by drawing KL, and in KL or KL produced ſind the
              <lb/>
            point M (by Lemma 4.) </s>
            <s xml:id="echoid-s527" xml:space="preserve">ſuch, that all the lines drawn from it cutting the given
              <lb/>
            circles ſhall cut off ſimilar ſegments; </s>
            <s xml:id="echoid-s528" xml:space="preserve">and let KL cut the circumferences, one
              <lb/>
            of them in the points A and I, and the other in the points E and H; </s>
            <s xml:id="echoid-s529" xml:space="preserve">and join-
              <lb/>
            ing MD, make it as MD: </s>
            <s xml:id="echoid-s530" xml:space="preserve">MA:</s>
            <s xml:id="echoid-s531" xml:space="preserve">: MH: </s>
            <s xml:id="echoid-s532" xml:space="preserve">MN. </s>
            <s xml:id="echoid-s533" xml:space="preserve">Then through the points D and
              <lb/>
            N draw a circle which ſhall alſo touch the given circle whoſe center is K, by
              <lb/>
            Prob. </s>
            <s xml:id="echoid-s534" xml:space="preserve">XII. </s>
            <s xml:id="echoid-s535" xml:space="preserve">and I ſay that this circle will alſo touch the other given circle whoſe
              <lb/>
            center is L. </s>
            <s xml:id="echoid-s536" xml:space="preserve">For let B be the point of contact and BM be drawn cutting the
              <lb/>
            circle K in C, and the circle L in F and G; </s>
            <s xml:id="echoid-s537" xml:space="preserve">then by Lemma 5. </s>
            <s xml:id="echoid-s538" xml:space="preserve">MB x MG </s>
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