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-s435" xml:space="preserve">
              <pb o="(10)" file="0022" n="22"/>
            let another circle be deſcribed, touching this laſt, and alſo the line XH and paſſ-
              <lb/>
            ing through the point A by Problem X, and I ſay that E the center of this circle
              <lb/>
            will alſo be the center of the circle required; </s>
            <s xml:id="echoid-s436" xml:space="preserve">as will appear by taking equals from
              <lb/>
            equals, or adding equals to equals, as the aſſigned Caſe and Data ſhall require.</s>
            <s xml:id="echoid-s437" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s438" xml:space="preserve">
              <emph style="sc">The</emph>
            Caſes are four, though Vieta makes but three.</s>
            <s xml:id="echoid-s439" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s440" xml:space="preserve">
              <emph style="sc">Case</emph>
            Iſt. </s>
            <s xml:id="echoid-s441" xml:space="preserve">If it be required that the circle ſhould touch both the others ex-
              <lb/>
            ternally then BG muſt be taken equal to the difference of the Semidiameters of
              <lb/>
            the two given circles, and ZX muſt be taken in BZ produced.</s>
            <s xml:id="echoid-s442" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s443" xml:space="preserve">
              <emph style="sc">Case</emph>
            2d. </s>
            <s xml:id="echoid-s444" xml:space="preserve">If it be required that the circle ſhall touch and include both the
              <lb/>
            given ones; </s>
            <s xml:id="echoid-s445" xml:space="preserve">then BG muſt be taken equal to the difference, as in Caſe 1ſt, but
              <lb/>
            ZX muſt be taken in BZ itſelf.</s>
            <s xml:id="echoid-s446" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s447" xml:space="preserve">
              <emph style="sc">Case</emph>
            3d. </s>
            <s xml:id="echoid-s448" xml:space="preserve">If it be required that the circle ſhould touch and include the greater
              <lb/>
            of the given circles, and touch externally the other whoſe center is B; </s>
            <s xml:id="echoid-s449" xml:space="preserve">then BG
              <lb/>
            muſt be taken equal to the ſum of the Radii of the given circles, and ZX muſt
              <lb/>
            be taken in BZ itſelf.</s>
            <s xml:id="echoid-s450" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s451" xml:space="preserve">
              <emph style="sc">Case</emph>
            4th. </s>
            <s xml:id="echoid-s452" xml:space="preserve">If it be required that the circle ſhould touch the greater of the
              <lb/>
            given circles externally, and touch and include the leſſer; </s>
            <s xml:id="echoid-s453" xml:space="preserve">then BG muſt be taken
              <lb/>
            equal to the ſum of the Radii, and ZX muſt be taken in BZ produced.</s>
            <s xml:id="echoid-s454" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div25" type="section" level="1" n="25">
          <head xml:id="echoid-head30" xml:space="preserve">PROBLEM XII .</head>
          <p>
            <s xml:id="echoid-s455" xml:space="preserve">
              <emph style="sc">Having</emph>
            two points given B and D, and like wiſe a circle whoſe center is A; </s>
            <s xml:id="echoid-s456" xml:space="preserve">to de-
              <lb/>
            ſcribe another circle which ſhall paſs through the given points, and touch the
              <lb/>
            given circle.</s>
            <s xml:id="echoid-s457" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s458" xml:space="preserve">
              <emph style="sc">Let</emph>
            DB be joined, as alſo AB, and let AB be produced to cut the given
              <lb/>
            circle in the points I and K, then let BH be taken a 4th proportional to DB,
              <lb/>
            BK, BI; </s>
            <s xml:id="echoid-s459" xml:space="preserve">ſo that BD X BH = BI X BK: </s>
            <s xml:id="echoid-s460" xml:space="preserve">from H let a Tangent HF be drawn
              <lb/>
            to the given circle; </s>
            <s xml:id="echoid-s461" xml:space="preserve">and BF be joined and cut the circle again in G: </s>
            <s xml:id="echoid-s462" xml:space="preserve">and let DG
              <lb/>
            be drawn cutting the given circle again in E, and laſtly through the points D,
              <lb/>
            B, G, let a circle be drawn, I ſay it will touch the given circle in G.</s>
            <s xml:id="echoid-s463" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s464" xml:space="preserve">
              <emph style="sc">For</emph>
            joining EF; </s>
            <s xml:id="echoid-s465" xml:space="preserve">becauſe the rectangle DBH = the Rectangle KBI, i.</s>
            <s xml:id="echoid-s466" xml:space="preserve">e. </s>
            <s xml:id="echoid-s467" xml:space="preserve">the
              <lb/>
            Rectangle GBF, thereforethefourpoints D, H, F, G, are in a circle; </s>
            <s xml:id="echoid-s468" xml:space="preserve">and hence the
              <lb/>
            angle HFB = the angle GDB: </s>
            <s xml:id="echoid-s469" xml:space="preserve">(for in the two firſt ſigures one is theexternal angle
              <lb/>
            of a quadrilateral figure, and the other is the internal and oppoſite, and in the two
              <lb/>
            laſt figures theſe angles are in the ſame ſegment.) </s>
            <s xml:id="echoid-s470" xml:space="preserve">But the angle HFB = the angle
              <lb/>
            GEF by Eu. </s>
            <s xml:id="echoid-s471" xml:space="preserve">III. </s>
            <s xml:id="echoid-s472" xml:space="preserve">32. </s>
            <s xml:id="echoid-s473" xml:space="preserve">hence GEF = GDB: </s>
            <s xml:id="echoid-s474" xml:space="preserve">therefore the triangles GEF and
              <lb/>
            GDB are ſimilar and under the ſame vertex, and therefore by Lemma 3. </s>
            <s xml:id="echoid-s475" xml:space="preserve">the cir-
              <lb/>
            cles deſcribed about them will touch each other in the common vertex G.</s>
            <s xml:id="echoid-s476" xml:space="preserve"/>
          </p>
          <note symbol="*" position="foot" xml:space="preserve">There are other Conſtructions of this Problem in Hugo de Omerique, Simpſon, and the Mathematician.
            <lb/>
          See alſo Monthly Review for Oct. 1764, where the Reviewer is pleaſed to ſpeak favourably of the 1ſt Edi-
            <lb/>
          tion of this work, but wiſhes that ſome modern ſolutions of theſe Problems had been inſerted, which, he
            <lb/>
          ſays, are more conciſe and elegant than any which are to be met with in the works of the Antients. The
            <lb/>
          Editor acknowledges that the conſtruction there given is more ſimple than Vieta’s; but Vieta is not an
            <lb/>
          Antient, and he knows of no others that exceed his.</note>
        </div>
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