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

34(22) biſecting planes; let this right line be EF. Moreover, the center of the
ſphere ſought will alſo be in a plane biſecting the inclination of the two planes
AH and GH, and the interſection of this laſt biſecting plane with the right
line EF will give a point D, which will be the center of the ſphere required.

PROBLEM V.

Having three planes AB, BC, CD, given, and alſo a point H; to find
a ſphere which ſhall paſs through the given point, and likewiſe touch the
three given planes.

Suppose it done. The three planes, by what was ſaid under the laſt pro-
poſition, will give a right line in poſition, in which will be the center of the
ſphere required. Let this right line be GE, perpendicular to which from H
the given point let HI be drawn, which therefore will be given in magnitude
and poſition. Let HI be produced, and FI taken equal to HI; the point
F will then be given. Now ſince the center of the ſphere required is in the
line GE, and FH is perpendicular thereto and biſected thereby, and one
extreme H is by hypotheſis in the ſurface of the ſaid ſphere, the other extreme
F will be ſo too. Nay even a circle deſcribed with I center and IH radius
in a plane perpendicular to GE will be in the ſaid ſpherical ſurface. Here
then we have a circle given in magnitude and poſition, and taking any one
of the given planes AB, by an evident corollary from Problem II. of this
Supplement, a ſphere may be deſcribed which will touch the given plane,
and likewiſe have the given circle in it’s ſurface; and ſuch a ſphere will
anſwer every thing here required.

PROBLEM VI.

Having three planes ED, DB, BC, given, and alſo a ſphere RM, to
conſtruct a ſphere which ſhall touch the given one, and likewiſe the three
given planes.

Suppose it done, and that the ſphere ERCA is the required one, viz.
touches the ſphere in R, and the planes in E, A, C. Let the center of this
ſphere be O; then drawing RO, EO, AO, CO, they will all be equal, and
RO will paſs through M the center of the given ſphere; and EO, AO, CO,
will be perpendicular to the planes ED, DB, BC. Let OU, OG, OI, be
made each equal to OM; and through the points U, G and I, let the planes
UP, GH, IN, be ſuppoſed drawn parallel to the given ones ED, DB, BC,
reſpectively. Since OR is equal to OE, and OM equal to OU, RM will

31 ((19))	32 ((20))
33 ((21))	34 ((22))
35 ((23))	36 ((24))
37 ((25))	38 ((26))
39 ((27))	40 ((28))

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