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

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32(20) diameter of the circle given in poſition, and therefore the points A and D
will alſo be given.
Letus now ſuppoſe the thing done, and that the center of the ſphere
ſought is E, which, as obſerved before, muſt be in the line CB.
Drawing
EF, EA, ED, theſe lines muſt be equal, ſince the points F, A, D, have
been ſhewn to be in the ſurface of the ſphere.
But theſe lines EF, EA, ED,
are in the ſame plane, ſince FB and AD are parallel, and BC perpendicular
to each of them.
If therefore a circle be deſcribed to paſs through the three
points F, A, D, whoſe center is E, it will be in the line CB, and will be
the center of the ſphere required.
PROBLEM II.
Having three points N, O, M, given, and a plane AD, to deſcribe a-
ſphere which ſhall paſs through the three given points;
and alſo touch the
given plane.
Let a circle ENOM be deſcribed to paſs through the three given points,
it will be in the ſurface of the ſphere ſought, from what was ſaid under the
former Problem.
From it’s center I let a perpendicular to it’s plane IA be
erected;
the center of the ſphere ſought will be in this line IA; let the line
IA meet the given plane in the point A, which point will be therefore given.
From the center of the given circle ENOM, let a perpendicular to the given
plane ID be drawn, the point D will then be given, and therefore the line
AD both in poſition and magnitude, as likewiſe the lines ID, IA, and the
plane of the triangle ADI.
But the plane of the circle NOM is alſo given
in poſition, and therefore alſo their common ſection EIF, and hence the
points E and F.
Suppoſe now the thing done, and that the center of the ſphere ſought is B.
Draw BE, BF, and BC parallel to ID. Since the triangle ADI, and the
line EIF are in the ſame plane, therefore BE, BF, BC, will be in the ſame
plane.
But the line ID is perpendicular to the given plane, therefore the
line BC parallel to it, will alſo be perpendicular to the given plane.
Since
then a ſphere is to be deſcribed to touch the plane AD, a perpendicular BC
from it’s center B will give the point of contact C;
and BC, BE, BF will be
equal, and it has been proved that they are in a plane given in poſition, in
which plane is alſo the right line AD.
The queſtion is then reduced to this,
Having two points E and F given, as alſo a right line AD in the ſame

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