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

Page concordance

< >
Scan Original
21 (9)
22 (10)
23 (11)
24 (12)
25 (13)
26 (14)
27 (15)
28 (16)
29 (17)
30
31 (19)
32 (20)
33 (21)
34 (22)
35 (23)
36 (24)
37 (25)
38 (26)
39 (27)
40 (28)
41 (29)
42
43
44
45
46
47
48
49
50
< >
page |< < ((17)) of 161 > >|
    <echo version="1.0RC">
      <text xml:lang="en" type="free">
        <div xml:id="echoid-div34" type="section" level="1" n="34">
          <pb o="(17)" file="0029" n="29"/>
          <p>
            <s xml:id="echoid-s673" xml:space="preserve">
              <emph style="sc">Case</emph>
            3d. </s>
            <s xml:id="echoid-s674" xml:space="preserve">Suppoſe the given point A to lie in the given circle, whoſe center
              <lb/>
            is B.</s>
            <s xml:id="echoid-s675" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s676" xml:space="preserve">
              <emph style="sc">Then</emph>
            joining AB and continuing it to meet the given circumference in C and
              <lb/>
            O, biſect AC in E, and from E towards O ſetting off EH = BC the given
              <lb/>
            Radius, and with A and B Foci and EH tranſverſe Axis deſcribing an Ellipſe
              <lb/>
            EKH, it will evidently be the Locus required.</s>
            <s xml:id="echoid-s677" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div35" type="section" level="1" n="35">
          <head xml:id="echoid-head41" xml:space="preserve">PROBLEM VI.</head>
          <p>
            <s xml:id="echoid-s678" xml:space="preserve">
              <emph style="sc">Having</emph>
            a right line BC given, and alſo a circle whoſe center is A, to deter-
              <lb/>
            mine the Locus of the centers of the circles which ſhall be touched both by the
              <lb/>
            given right line and alſo by the given circle.</s>
            <s xml:id="echoid-s679" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s680" xml:space="preserve">
              <emph style="sc">There</emph>
            are three Caſes, but they are all comprchended under one general
              <lb/>
            ſolution.</s>
            <s xml:id="echoid-s681" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s682" xml:space="preserve">
              <emph style="sc">Case</emph>
            1ſt. </s>
            <s xml:id="echoid-s683" xml:space="preserve">Let the given right line be without the given circle, and let it be
              <lb/>
            required that the circles to be deſcribed be touched outwardly by the given circle.</s>
            <s xml:id="echoid-s684" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s685" xml:space="preserve">
              <emph style="sc">Case</emph>
            2d. </s>
            <s xml:id="echoid-s686" xml:space="preserve">Let the given right line be without the given circle, and let it be
              <lb/>
            required that the circles to be deſcribed, be touched inwardly by the given
              <lb/>
            circle.</s>
            <s xml:id="echoid-s687" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s688" xml:space="preserve">
              <emph style="sc">Case</emph>
            3d. </s>
            <s xml:id="echoid-s689" xml:space="preserve">Let the given right line be within the given circle, and then the
              <lb/>
            circles to be deſcribed muſt be touched outwardly by the given circle.</s>
            <s xml:id="echoid-s690" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div36" type="section" level="1" n="36">
          <head xml:id="echoid-head42" xml:space="preserve">
            <emph style="sc">General</emph>
            <emph style="sc">Solution.</emph>
          </head>
          <p>
            <s xml:id="echoid-s691" xml:space="preserve">
              <emph style="sc">From</emph>
            the given center A let fall a perpendicular AG to the given line BC,
              <lb/>
            which meets the given circumference in D [or in Caſes 2d and 3d is produced
              <lb/>
            to meet it in D] and biſecting DG in F, and ſetting off FM = FA (which is the
              <lb/>
            ſame thing as making GM = AD the given Radius) and through M drawing
              <lb/>
            MLK parallel to the given line BC, with A Focus and LK Directrix deſcribe a
              <lb/>
            Parabola, and it will be the Locus of the centers of the circles required; </s>
            <s xml:id="echoid-s692" xml:space="preserve">for
              <lb/>
            from the property of the Curve FA = FM, and adding equals to equals, or
              <lb/>
            ſubtracting equals from equals, as the Caſe requires, FD = FG.</s>
            <s xml:id="echoid-s693" xml:space="preserve"/>
          </p>
        </div>
      </text>
    </echo>
Impressum