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equal to UE: but RM is given in magnitude, being the radius of the given
<br />
ſphere, therefore UE is alſo given in magnitude. And ſince OE is perpen-
<br />
dicular to the plane DE, it will be alſo to the plane PU which is parallel
<br />
thereto. UE then being given in magnitude, and being the interval be-
<br />
tween two parallel planes DE, PU, whereof DE is given in poſition by hypo-
<br />
theſis, the other PU will alſo be given in poſition. In the ſame manner it
<br />
may be proved that the planes GH, IN, are given in poſition, and that the
<br />
lines OG, OI, are perpendicular thereto reſpectively, and each alſo equal to
<br />
OM. A ſphere therefore deſcribed with center O and OM radius will touch
<br />
the three planes PU, GH, IN, given in poſition: but the point M is given,
<br />
being the center of the given ſphere. The queſtion is then reduced to this,
<br />
Having three planes given PU, GH, IN, and a point M, to find the radius
<br />
of a ſphere which ſhall touch the given planes, and paſs through the given
<br />
point; which is the ſame as the preceeding Problem. [And this radius be-
<br />
ing increaſed or diminiſhed by MR, according as R is taken in the further or
<br />
nearer ſurface of the given ſphere, will give the radius of a ſphere which will
<br />
touch the three given planes DE, DB, BC, and likewiſe the given
<br />
ſphere. ]
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<p xmlns="http://www.w3.org/1999/xhtml">
<span class="caps" style="font-size: 75%">By</span>
a like method, when among the Data there are no points, but only
<br />
planes and ſpheres, we ſhall always be able to ſubſtitute a given point in the
<br />
place of a given ſphere.
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<span class="headingNumber">43.</span>
<span class="head">PROBLEM VII.</span>
</h2>
<p xmlns="http://www.w3.org/1999/xhtml">
<span class="caps" style="font-size: 75%">Having</span>
two points H, M, as alſo two planes AB, BC, given, to find a
<br />
ſphere which ſhall paſs through the given points, and touch the given
<br />
planes.
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<p xmlns="http://www.w3.org/1999/xhtml">
<span class="caps" style="font-size: 75%">Draw</span>
HM and biſect it in I, the point I will be given, through the
<br />
point I let a plane be erected perpendicular to the right line HM, this plane
<br />
will be given in poſition, and the center of the ſphere required will be in this
<br />
plane. But becauſe it is alſo to touch the planes AB, BC, its center will be
<br />
alſo in another plane given in poſition (by what has been proved, Prob. IV.) and therefore in a right line which is their interſection, given in poſition,
<br />
which let be GE; to which line GE from one of the given points M demit-
<br />
ting a perpendicular MF, it will be given in magnitude and poſition, and
<br />
being continued to D ſo that FD equals MF, the point D will be given; and, from what has been proved before, will be in the ſpherical ſurface.
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