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

Table of contents

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[11.] PROBLEM IV.
Page: 15 ((3))
[12.] PROBLEM V.
Page: 16 ((4))
[13.] The general Solution.
Page: 16 ((4))
[14.] PROBLEM VI.
Page: 17 ((5))
[15.] The general Solution.
Page: 17 ((5))
[16.] PROBLEM VII.
Page: 18 ((6))
[17.] LEMMA I.
Page: 19 ((7))
[18.] PROBLEM VIII.
Page: 19 ((7))
[19.] Mr. Simpſon conſtructs the Problem thus.
Page: 19 ((7))
[20.] PROBLEM IX.
Page: 19 ((7))
[21.] LEMMA II.
Page: 20 ((8))
[22.] LEMMA III.
Page: 20 ((8))
[23.] PROBLEM X.
Page: 20 ((8))
[24.] PROBLEM XI.
Page: 21 ((9))
[25.] PROBLEM XII .
Page: 22 ((10))
[26.] LEMMA IV.
Page: 23 ((11))
[27.] LEMMA V.
Page: 23 ((11))
[28.] PROBLEM XIII.
Page: 23 ((11))
[29.] PROBLEM XIV.
Page: 24 ((12))
[30.] SUPPLEMENT. PROBLEM I.
Page: 26 ((14))
[31.] PROBLEM II.
Page: 26 ((14))
[32.] PROBLEM III.
Page: 26 ((14))
[33.] PROBLEM IV.
Page: 28 ((16))
[34.] PROBLEM V.
Page: 28 ((16))
[35.] PROBLEM VI.
Page: 29 ((17))
[36.] General Solution.
Page: 29 ((17))
[37.] A SECOND SUPPLEMENT, BEING Monſ. DE FERMAT’S Treatiſe on Spherical Tangencies. PROBLEM I.
Page: 31 ((19))
[38.] PROBLEM II.
Page: 32 ((20))
[39.] PROBLEM III.
Page: 33 ((21))
[40.] PROBLEM IV.
Page: 33 ((21))
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page |< < ((19)) of 161 > >|
31(19)
A SECOND
SUPPLEMENT,
BEING
Monſ. DE FERMAT’S Treatiſe on
Spherical Tangencies.
PROBLEM I.
HAVING four points N, O, M, F, given, to deſcribe a ſphere which
ſhall paſs through them all.
Taking any three of them N, O, M, ad libitum, they will form a triangle,
about which a circle ANOM may be circumſcribed, which will be given in
magnitude and poſition. That this circle is in the ſurface of the ſphere
ſought appears from hence; becauſe if a ſphere be cut by any plane, the
ſection will be a circle; but only one circle can be drawn to paſs through the
three given points N, O, M; therefore this circle muſt be in the ſurface of
the ſphere. Let the center of this circle be C, from whence let CB be
erected perpendicular to it’s plane; it is evident that the center of the ſphere
ſought will be in this line CB. From the fourth given point F let FB be
drawn perpendicular to CB, which FB will be alſo given in magnitude and
poſition. Through C draw ACD parallel to FB, and this line will be

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