Full text: Pergaeus, Apollonius: The two books of Apollonius Pergaeus, concerning tangencies, as they have been restored by Franciscus Vieta and Marinus Ghetaldus

to find the center of a circle which will paſs through the two points, and like-
wiſe touch the right line, which is the VIIth of the preceeding Problems.

39. PROBLEM III.

Having three points N, O, M, given, as likewiſe a ſphere IG, to de-
ſcribe a ſphere which will paſs through the three given points, and likewiſe
touch the given ſphere.

The circle NOM in the ſurface of the ſphere ſought is given, and a per-
pendicular to its plane from it’s center FA being drawn, the center of the
ſphere required will be in this line. From I the center of the given ſphere
let IB be drawn perpendicular to FA, and through F, ED parallel to IB,
which, from what has been before proved, will be in the plane of the circle
NOM, and the points E and D will be given.

Suppoſe now the thing done, and that the center of the ſphere required is
C. Then the lines CI, CE, CD, will be in the ſame plane, which is given, as
the points I, E, and D are given. But the point of contact of two ſpheres is
in the line joining their centers; therefore the ſphere ſought will touch the
ſphere given in the point G, and the line IC will exceed the lines EC, ED, by
IG the radius of the given ſphere: with center I therefore and this diſtance
IG let a circle be deſcribed in the plane of the lines CI, CE, CD, and it
will paſs through the point G and be given in magnitude and poſition; but
the points D and E are alſo in the ſame plane; and therefore the queſtion is
reduced to this, Having two points E and D given, as likewiſe a circle
IGH, to find the center of a circle which will paſs through the two points
and likewiſe touch the circle, which is the XIIth of the preceeding Problems.

40. PROBLEM IV.

Having four planes AH, AB, BC, HG, given; it is required to de-
ſcribe a ſphere which ſhall touch them all four.

If two planes touch a ſphere, the center of that ſphere will be in a plane
beſecting the inclination of the other two. And if the planes be parallel, it
will be in a parallel plane beſecting their interval. This being allowed,
which is too evident to need further proof; the center of the ſphere ſought
will be in a plane biſecting the inclination of two planes CB and BA; it will
likewiſe be in another plane biſecting the inclination of the two planes BA and
AH; and therefore in a right line, which is the common ſection of theſe two

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