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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          <head xml:id="echoid-head21" xml:space="preserve">PROBLEM VII.</head>
          <p>
            <s xml:id="echoid-s268" xml:space="preserve">
              <emph style="sc">Having</emph>
            two points A and B given in poſition, and likewiſe a right line EF
              <lb/>
            given in poſition, it is required to find the center of a circle, which ſhall paſs
              <lb/>
            through the given points and touch the given line.</s>
            <s xml:id="echoid-s269" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s270" xml:space="preserve">
              <emph style="sc">Case</emph>
            Iſt. </s>
            <s xml:id="echoid-s271" xml:space="preserve">When the points A and B are joined, ſuppoſe AB to be parallel to
              <lb/>
            EF: </s>
            <s xml:id="echoid-s272" xml:space="preserve">then biſecting AB in D, and through D drawing DC perpendicular to it,
              <lb/>
            DC will alſo be perpendicular to EF: </s>
            <s xml:id="echoid-s273" xml:space="preserve">draw a circle therefore which will paſs
              <lb/>
            through the three points A, B, andC, (by Euc. </s>
            <s xml:id="echoid-s274" xml:space="preserve">IV. </s>
            <s xml:id="echoid-s275" xml:space="preserve">5.) </s>
            <s xml:id="echoid-s276" xml:space="preserve">and it will be the circle
              <lb/>
            required: </s>
            <s xml:id="echoid-s277" xml:space="preserve">(by a Corollary from Euc. </s>
            <s xml:id="echoid-s278" xml:space="preserve">III. </s>
            <s xml:id="echoid-s279" xml:space="preserve">1. </s>
            <s xml:id="echoid-s280" xml:space="preserve">and another from Euc.
              <lb/>
            </s>
            <s xml:id="echoid-s281" xml:space="preserve">III. </s>
            <s xml:id="echoid-s282" xml:space="preserve">16.)</s>
            <s xml:id="echoid-s283" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s284" xml:space="preserve">
              <emph style="sc">Case</emph>
            2d. </s>
            <s xml:id="echoid-s285" xml:space="preserve">Suppoſe AB not parallel to EF, but being produced meets it in
              <lb/>
            E: </s>
            <s xml:id="echoid-s286" xml:space="preserve">then from EF take the line EC ſuch, that its ſquare may be equal to the
              <lb/>
            rectangle BEA, and through the points A, B, C, deſcribe a circle, and it will be
              <lb/>
            the circle required by Euc. </s>
            <s xml:id="echoid-s287" xml:space="preserve">III. </s>
            <s xml:id="echoid-s288" xml:space="preserve">37.</s>
            <s xml:id="echoid-s289" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s290" xml:space="preserve">
              <emph style="sc">This</emph>
            is Vieta’s Solution. </s>
            <s xml:id="echoid-s291" xml:space="preserve">But Mr. </s>
            <s xml:id="echoid-s292" xml:space="preserve">Thomas Simpſon having conſtructed this,
              <lb/>
            and ſome of the following, both in the Collection of Problems at the end of his
              <lb/>
            Algebra, and alſo among thoſe at the end of his Elements of Geometry, I ſhall
              <lb/>
            add one of his Conſtructions.</s>
            <s xml:id="echoid-s293" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s294" xml:space="preserve">
              <emph style="sc">Let</emph>
            A and B be the points given, and CD the given line: </s>
            <s xml:id="echoid-s295" xml:space="preserve">drawing AB and
              <lb/>
            biſecting it in F, through E let EF be drawn perpendicular to AB and meeting
              <lb/>
            CD in F: </s>
            <s xml:id="echoid-s296" xml:space="preserve">and from any point H in EF draw HG perpendicular to CD, and
              <lb/>
            having drawn BF, to the ſame apply HI = HG, and parallel thereto draw BK
              <lb/>
            meeting EF in K: </s>
            <s xml:id="echoid-s297" xml:space="preserve">then with center K and radius BK let a circle be deſcribed,
              <lb/>
            and the thing is done: </s>
            <s xml:id="echoid-s298" xml:space="preserve">join KA, and draw KL perpendicular to CD, then be-
              <lb/>
            cauſe of the parallel lines, HG: </s>
            <s xml:id="echoid-s299" xml:space="preserve">HI:</s>
            <s xml:id="echoid-s300" xml:space="preserve">: KL: </s>
            <s xml:id="echoid-s301" xml:space="preserve">KB; </s>
            <s xml:id="echoid-s302" xml:space="preserve">whence as HG and HI are
              <lb/>
            equal, KL and KB are likewiſe equal. </s>
            <s xml:id="echoid-s303" xml:space="preserve">But it is evident from the Conſtruction
              <lb/>
            that KA = KB, therefore KB = KL = KA.</s>
            <s xml:id="echoid-s304" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s305" xml:space="preserve">
              <emph style="sc">Because</emph>
            two equal lines HI and Hi may be applied from H to BF each
              <lb/>
            equal to HG, the Problem will therefore admit of two Solutions, as the Figure
              <lb/>
            ſhews: </s>
            <s xml:id="echoid-s306" xml:space="preserve">except in the caſe when one of the given points, A for inſtance, is given
              <lb/>
            in the line CD, for then the Problem becomes more ſimple, and admits but of
              <lb/>
            one conſtruction, as the center of the circle required muſt be in the line EF con-
              <lb/>
            tinued, as alſo in the perpendicular raiſed from A to CD, and therefore in their
              <lb/>
            common interſection: </s>
            <s xml:id="echoid-s307" xml:space="preserve">and this is the limit of poſſibility; </s>
            <s xml:id="echoid-s308" xml:space="preserve">for ſhould the line CD
              <lb/>
            paſs between the given points, the Problem is impoſſible.</s>
            <s xml:id="echoid-s309" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s310" xml:space="preserve">N. </s>
            <s xml:id="echoid-s311" xml:space="preserve">B. </s>
            <s xml:id="echoid-s312" xml:space="preserve">
              <emph style="sc">Tho</emph>
            ’ Vieta does not take notice that this Problem is capable of two
              <lb/>
            anſwers, yet this is as evident from his conſtruction, as from Mr. </s>
            <s xml:id="echoid-s313" xml:space="preserve">Simpſon’s, for
              <lb/>
            EC (the mean proportional between E B and EA) may be ſet off upon the given
              <lb/>
            line EF either way from the given point E.</s>
            <s xml:id="echoid-s314" xml:space="preserve"/>
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