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

Thercfore there are given three points H, M, D, as likewiſe a plane AB,
or AC, through which points the ſphere is to paſs, and alſo touch the given
plane. Hence it appears that this Problem is reduced to the IId of this

Before we proceed, the following eaſy Lemmas muſt be premiſed.

44. LEMMA I.

Let there be a circle BCD, and a point E taken without it, and iſ from
E a line EDOB be drawn to paſs through the center, and another line ECA
to cut it any ways; we know from the Elements that the rectangle AEC is
equal to the rectangle BED. Let us now ſuppoſe a ſphere whoſe center is O,
and one of whoſe great circles is ACDB; if from the ſame point E a line
ECA be any-how drawn to meet the ſpherical ſurface in the points C and A,
I ſay the rectangle AEC will ſtill be equal to the rectangle BED. For if we
ſuppoſe the circle and right line ECA to revolve upon EDB as an immove-
able axis, the lines EC and EA will not be changed, becauſe the points C
and A deſcribe circles whoſe planes are perpendicular to that axis; and
therefore the rectangle AEC will in any plane be ſtill equal to the rectangle


By the ſame method of reaſoning, the Vth Lemma immediately preceed-
ing Problem XIII, in the Treatiſe of Circular Tangencies, may be extended
alſo to ſpheres, viz. that in any plane (ſee the Figures belonging to that
Lemma) MG X MB = MH X MA. And alſo that MF X MC = ME X MI.


Let there be two ſpheres YN, XM, through whoſe centers let the right
line RYNXMU paſs, and let it be as the radius YN to the radius XM, ſo
YU to XU; and from the point U let a line UTS be drawn in any plane,
and let the rectangle S U T be equal to the rectangle RUM; I fay that if
any ſphere OTS be deſcribed to paſs through the points T and S, and to
touch one of the given ſpheres XM as in O, it will alſo touch the other
given ſphere YN. For joining UO, and producing it to meet the ſurface of
the ſphere OTS in Q; the rectangle QUO = the rectangle SUT, by
Lemma I. but the rectangle SUT = the rectangle RUM. by conſtruction,

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