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to find the center of a circle which will paſs through the two points, and like-
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wiſe touch the right line, which is the VIIth of the preceeding Problems.
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<span class="headingNumber">39.</span>
<span class="head">PROBLEM III.</span>
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<span class="caps" style="font-size: 75%">Having</span>
three points N, O, M, given, as likewiſe a ſphere IG, to de-
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ſcribe a ſphere which will paſs through the three given points, and likewiſe
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touch the given ſphere.
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The circle NOM in the ſurface of the ſphere ſought is given, and a per-
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pendicular to its plane from it’s center FA being drawn, the center of the
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ſphere required will be in this line. From I the center of the given ſphere
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let IB be drawn perpendicular to FA, and through F, ED parallel to IB,
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which, from what has been before proved, will be in the plane of the circle
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NOM, and the points E and D will be given.
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Suppoſe now the thing done, and that the center of the ſphere required is
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C. Then the lines CI, CE, CD, will be in the ſame plane, which is given, as
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the points I, E, and D are given. But the point of contact of two ſpheres is
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in the line joining their centers; therefore the ſphere ſought will touch the
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ſphere given in the point G, and the line IC will exceed the lines EC, ED, by
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IG the radius of the given ſphere: with center I therefore and this diſtance
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IG let a circle be deſcribed in the plane of the lines CI, CE, CD, and it
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will paſs through the point G and be given in magnitude and poſition; but
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the points D and E are alſo in the ſame plane; and therefore the queſtion is
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reduced to this, Having two points E and D given, as likewiſe a circle
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IGH, to find the center of a circle which will paſs through the two points
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and likewiſe touch the circle, which is the XIIth of the preceeding Problems.
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<span class="headingNumber">40.</span>
<span class="head">PROBLEM IV.</span>
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<span class="caps" style="font-size: 75%">Having</span>
four planes AH, AB, BC, HG, given; it is required to de-
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ſcribe a ſphere which ſhall touch them all four.
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<span class="caps" style="font-size: 75%">If</span>
two planes touch a ſphere, the center of that ſphere will be in a plane
<br />
beſecting the inclination of the other two. And if the planes be parallel, it
<br />
will be in a parallel plane beſecting their interval. This being allowed,
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which is too evident to need further proof; the center of the ſphere ſought
<br />
will be in a plane biſecting the inclination of two planes CB and BA; it will
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likewiſe be in another plane biſecting the inclination of the two planes BA and
<br />
AH; and therefore in a right line, which is the common ſection of theſe two
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