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

26. LEMMA IV.

Having two circles ABCI and EFGH given, it is required to find a point
M, in the line joining their centers, or in that line continued, ſuch, that any
line drawn through the ſaid point M, cutting both the circles, ſhall always cut
off ſimilar ſegments.

Let the line KL joining the centers be ſo cut or produced, that KM may be
to LM in the given ratio of the Radii AK to EL [in the caſe of producing
KL we muſt make it as AK-EL: EL: : KL: LM, for then by compoſition it
will be AK: EL: : KM: LM] and then I ſay that the point M will be the
point required. For from it drawing any line MGFCB cutting the circle
ABCI in B and C, and the circle EFGH in F and G; and let B and F be the
correſpondent points moſt diſtant from M, and C and G the correſpondent
points that are nearer; and let be joined BK, CK, FL, GL, and thereby two
triangles will be formed BKC, FLG. Now becauſe by Conſtri [?] ction KM: LM
: : KB: LF, KB and LF will be parallel by Euc. VI. 7. and for the ſame rea-
ſon KC and LG will be parallel; and therefore the triangles BKC and FLG
will be equiangled, and hence the ſegment BC will be ſimilar to the ſegment FG.

27. LEMMA V.

The point M being found as in the preceding Lemma, I ſay that it is a pro-
perty of the ſaid point, that MG x MB = MH x MA: as alſo that MF x
MC = ME x MI.

For joining CI and GH, it is evident that theſe lines will alſo be parallel. Hence MI: MC: : MH: MG, but MI: MC: : MB: MA, therefore
MH: MG: : MB: MA, and MG x MB = MH x MA. Again MH: MG
: : MF: ME, but MH: MG: : MI: MC, therefore MF: ME: : MI: MC
and MF x MC = ME x MI.

28. PROBLEM XIII.

Having two circles given in magnitude and poſition, whoſe centers are K
and L, and alſo a point D; to draw a circle which ſhall touch the two given
ones, and alſo paſs through the point D.

Join the given centers by drawing KL, and in KL or KL produced ſind the
point M (by Lemma 4.) ſuch, that all the lines drawn from it cutting the given
circles ſhall cut off ſimilar ſegments; and let KL cut the circumferences, one
of them in the points A and I, and the other in the points E and H; and join-
ing MD, make it as MD: MA: : MH: MN. Then through the points D and
N draw a circle which ſhall alſo touch the given circle whoſe center is K, by
Prob. XII. and I ſay that this circle will alſo touch the other given circle whoſe
center is L. For let B be the point of contact and BM be drawn cutting the
circle K in C, and the circle L in F and G; then by Lemma 5. MB x MG =

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