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

From A draw two perpendiculars to the right lines DB, ZC; viz. ADF and
AZX; and in theſe perpendiculars take DF, ZX, on “either ſide of D and Z,
equal to the Radius of the given circle: and through F [?] and X draw lines pa-
rallel to DB, ZC; viz. FG, XH; and then by the preceding Problem draw
a circle which ſhall paſs through the given center A and touch the two lines
FG, XH; and E the center of this circle will alſo be the center of the circle
required, as appears by ſubtracting equals from equals in Figure 1: and by adding
equals to equals in Figure 2.


If the two circles CEB and CED cut one another C, then I ſay a line drawn
from the point of ſection CBD cutting both circles, will cut off diſſimilar ſeg-
ments from thoſe circles.

1ſt. Suppose CB to be the Diameter of one of them: then draw to the
other point of ſection E the line CE, and joining EB, ED, the angle CEB will
be a right one, and the angle CED either greater or leſs than a right one, and
conſequently CD cannot be a Diameter of the other.

2dly. Suppose CBD not to paſs through the center of either: then through
C draw a Diameter CAG, and continue it to meet the other circle in F, and
join BG, DF: then the angle CBG is a right one, and the angle CDF is either
greater or leſs than a right one: and therefore the lines BG and DF are not
parallel: let H be the center of the other circle, and let a Diameter CHI be
drawn: draw DI and continue it meet to meet CG in K: then DIK will be pa-
rallel to BG: hence CB: CD: : CG: CK. But CI and CK are unequal,
(being both applied from the ſame point in a right angle) and therefore it cannot
be 2s CB: CD: : CG: CI: and hence it appears that the Segments CB and
CD are diſſimilar.


If through the legs of any triangle EDF (ſee Figure to Problem 10.) a line
BI be [?] drawn parallel to the baſe DF, ſo that there be conſtituted two ſimilar
triangles about the ſame vertex; and a circle be circumſcribed about each of
theſe triangles; theſe circles will touch one another in the common vertex E.

It is plain that they will either touch or cut each other in the point E: if
they cut each other, then by the preceding Lemma the Segments BE and DE
would be diſſimilar; but they are ſimilar, and they muſt therefore touch each


Having a point A, and alſo a right line BC, given in poſition; together
with a circle whoſe center is G given both in m [...] de and [?] poſition; to deſcribe


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.

Please use one of the following browsers to display this page correctly.

Thank you.

powered by Goobi viewer