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-s397" xml:space="preserve">
              <pb o="(9)" file="0021" n="21"/>
            a circle which ſhall paſs through the given point, and likewiſe touch both the
              <lb/>
            given line and the given circle.</s>
            <s xml:id="echoid-s398" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s399" xml:space="preserve">
              <emph style="sc">The</emph>
            given right line may either 1ſt cut, touch, or be entirely without the
              <lb/>
            given circle, and the given point be without the ſame or in the circum-
              <lb/>
            ference; </s>
            <s xml:id="echoid-s400" xml:space="preserve">or 2dly, it may cut the given circle, and the given point be within
              <lb/>
            the ſame, or in the circumference.</s>
            <s xml:id="echoid-s401" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s402" xml:space="preserve">
              <emph style="sc">Case</emph>
            Iſt. </s>
            <s xml:id="echoid-s403" xml:space="preserve">Suppoſe the given line BC to cut, touch, or fall entirely with-
              <lb/>
            out the given circle; </s>
            <s xml:id="echoid-s404" xml:space="preserve">and the given point A to be without the ſame, or in the
              <lb/>
            circumference: </s>
            <s xml:id="echoid-s405" xml:space="preserve">through G the center of the given circle draw DGFC per-
              <lb/>
            pendicular to BC, and joining DA, take DH a 4th proportional to DA, DC,
              <lb/>
            DF, ſo that DA X DH = DC X DF: </s>
            <s xml:id="echoid-s406" xml:space="preserve">then through the points A and H draw
              <lb/>
            a circle touching the line CB by Problem VII, and I ſay it will alſo touch the
              <lb/>
            given circle.</s>
            <s xml:id="echoid-s407" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s408" xml:space="preserve">Draw DB cutting the given circle in E, and join FE. </s>
            <s xml:id="echoid-s409" xml:space="preserve">Now becauſe the
              <lb/>
            triangles DEF, DCB are ſimilar, DF: </s>
            <s xml:id="echoid-s410" xml:space="preserve">DE:</s>
            <s xml:id="echoid-s411" xml:space="preserve">: DB: </s>
            <s xml:id="echoid-s412" xml:space="preserve">DC, and therefore DC
              <lb/>
            X DF = DB X DE. </s>
            <s xml:id="echoid-s413" xml:space="preserve">But DC X DF = DA X DH by Conſtruction. </s>
            <s xml:id="echoid-s414" xml:space="preserve">Hence
              <lb/>
            DB X DE = DA X DH, and therefore the points B, E, H, A, will be alſo in
              <lb/>
            a circle: </s>
            <s xml:id="echoid-s415" xml:space="preserve">but the point E is alſo in the given circle; </s>
            <s xml:id="echoid-s416" xml:space="preserve">therefore theſe circles either
              <lb/>
            touch or cut one another in that point. </s>
            <s xml:id="echoid-s417" xml:space="preserve">Let now BI be drawn from the point
              <lb/>
            of contact B perpendicular to the touching line BC to meet the circumference
              <lb/>
            again in I, and it will be a Diameter: </s>
            <s xml:id="echoid-s418" xml:space="preserve">and let EI be joined: </s>
            <s xml:id="echoid-s419" xml:space="preserve">then becauſe the
              <lb/>
            angles FED and BEI are vertical and each of them right ones, FEI will be a
              <lb/>
            continued ſtraight line: </s>
            <s xml:id="echoid-s420" xml:space="preserve">and it appears that the two circles will touch each
              <lb/>
            other by the preceding Lemmas.</s>
            <s xml:id="echoid-s421" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s422" xml:space="preserve">
              <emph style="sc">Case</emph>
            2d. </s>
            <s xml:id="echoid-s423" xml:space="preserve">Suppoſe the given line BC to cut the given circle, and the given
              <lb/>
            point to be within the ſame, or in the circumference; </s>
            <s xml:id="echoid-s424" xml:space="preserve">the Conſtruction and De-
              <lb/>
            monſtration are exactly the ſame as before, except that the angles FED and
              <lb/>
            BEI are not vertical but coincident, and ſo EI is coincident with EF.</s>
            <s xml:id="echoid-s425" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s426" xml:space="preserve">N.</s>
            <s xml:id="echoid-s427" xml:space="preserve">B. </s>
            <s xml:id="echoid-s428" xml:space="preserve">In either of theſe caſes if the point A coincide with E or be given in
              <lb/>
            the circumference, draw DEB, and erect BI perpendicular to CB to meet FE in
              <lb/>
            I, then upon BI as diameter deſcribe a circle, and the thing will manifeſtly be done.</s>
            <s xml:id="echoid-s429" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div24" type="section" level="1" n="24">
          <head xml:id="echoid-head29" xml:space="preserve">PROBLEM XI.</head>
          <p>
            <s xml:id="echoid-s430" xml:space="preserve">
              <emph style="sc">Having</emph>
            two circles given in magnitude and poſition, whoſe centers are A
              <lb/>
            and B, as likewiſe a right line CZ; </s>
            <s xml:id="echoid-s431" xml:space="preserve">to draw a circle which ſhall touch all three.</s>
            <s xml:id="echoid-s432" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s433" xml:space="preserve">
              <emph style="sc">From</emph>
            the center of the leſſer circle B let BZ be drawn perpendicular to CZ,
              <lb/>
            and in BZ (or in BZ continued as the caſe requires) let be taken ZX = AL the
              <lb/>
            Radius of the other circle; </s>
            <s xml:id="echoid-s434" xml:space="preserve">and through X let XH be drawn parallel to CZ,
              <lb/>
            and with center B and Radius BG, equal to the difference ( or ſum as the Caſe
              <lb/>
            requires) of the Radii of the two given circles, let a circle be deſcribed; </s>
            <s xml:id="echoid-s435" xml:space="preserve">and </s>
          </p>
        </div>
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