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

greater of the given circles the line EH = the difference of the Radii of the given
circles, and with A and B as Foci and EH Tranverſe Axis, let two oppoſite
Hyperbolas be deſcribed KEK and LHL: then I ſay that KEK will be the
locus of the centers of all the circles which can be drawn ſo as to be touched
outwardly by both the given circles; and LHL will be the locus of the centers
of all the circles which can be drawn ſo as to be touched inwardly by both the
given circles. For taking any point K in the Hyperbola KEK, and drawing KA
and KB, let theſe lines cut the given circumferences in F and G reſpectively: and make KQ = KA: then from the nature of the curve QB = EH, but by con-
ſtruction EH = BG - AF, therefore QB = BG - AF and hence QG = AF, and
then taking equals from equals KG = KF, which is a demonſtration of the 1ſt
Caſe.

Then with regard to the 2d Caſe, taking any point L in the Hyperbola
LHL, and drawing LB and LA and producing them to meet the concave cir-
cumferences in M and N, let alſo LR be taken equal to LB; then from the pro-
perty of the curve AR = EH, but EH (by conſtruction) = BM - NA; there-
fore AR = BM - NA, and NR = BM, and then adding equals to equals LN =
LM, which is a demonſtration of the 2d Caſe.

Case 3d. Suppoſe it be required that the circles to be deſcribed be touched
outwardly by one of the given circles, and inwardly by the other.

Then drawing AB, let it cut the convex circumferences in C and D, and the
concave ones in P and O, biſect PD in E, and from E towards B ſet off EH =
the ſum of the given radii. Then with A and B foci and EH tranſverſe axis,
let two oppoſite hyperbolas be deſcribed KEK and LHL: and KEK will be the
locus of the centers of the circles which are touched inwardly by the circle A
and outwardly by the circle B; and LHL will be the locus of the centers of
thoſe circles which are touched inwardly by B and outwardly by A. The demon-
ſtration mutatis mutandis is the ſame as before.

Case 4th. Suppoſe the given circle A to include B, and it be required that
the circles to be deſcribed be touched outwardly by them both.

Let AB cut the circumferences in C and D, P and O: and biſecting CD in
I, and ſetting off from I towards P, IL = the ſum of the ſemidiameters of the
given circles, and with A and B foci, and IL tranſverſe axis, deſcribing an
ellipſe LKI, it will be the locus of the centers of the circles required. For
taking any point K in the ellipſe, and drawing AK and BK, let AK be con-
tinued to meet one of the given circumferences in G, and let BK meet the other
F. Then from the property of the curve AK + BK = IL = AG + BF (by con- ### Note to user

Dear user,

In response to current developments in the web technology used by the Goobi viewer, the software no longer supports your browser.