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

MA x MH: but MA x MH = MD x MN by Conſtruction; therefore MB x
MG = MD x MN, and the points B, G, N, D, are in a circle. But the point
G is alſo in the circle L; therefore theſe circles either touch or cut each other in
the point G. Now the circles BND and BCI touch one another in B by con-
ſtruction; therefore the ſegment BC is ſimilar to the ſegment BG; and alſo by
conſtruction the ſegment BC is ſimilar to the ſegment FG; and therefore the
ſegment FG is ſimilar to the ſegment BG; and hence the circles FGE and BGD
touch one another in the point G.

The Caſes are three.

Case Iſt. Iſ the circle be required to touch and include both the given ones; then M muſt be taken in KL produced; and MN muſt be taken a fourth pro-
portional to MD, MA, MH, A being the moſt diſtant point of interſection in
the circle K, and H the neareſt point of interſection in the circle L.

Case 2d. If the circle be required to touch both the given ones externally; then alſo M muſt be taken in KL produced; and MN taken a fourth proportional
to MD, MA, MH, A being the neareſt point of interſection in the circle K,
and H the moſt diſtant in the circle L.

Case 3d. If the circle be required to touch and include the circle K, and to
touch L externally; then M muſt be taken in KL itſelf; and MN a fourth pro-
portional to MD, MA, MH, A being the moſt diſtant point in K, and H the
neareſt in L.

29. PROBLEM XIV.

Having three circles given whoſe centers are A, B, and D; to draw a fourth
which ſhall touch all three.

Let that whoſe center is A be called the 1ſt, that whoſe center is B the 2d,
and that whoſe center is D the 3d. Then with center B, and radius equal to
the differcnce, or ſum, as the caſe requires, of the ſemidiameters of the 1ſt and
2d circles, let an auxiliary circle be deſcribed; and likewiſe with D center, and
radius equal to the difference, or ſum, as the caſe requires, of the ſemidiameters
of the 1ſt and 3d circles, let another auxiliary circle be deſcribed; and laſtly by
the preceding Problem draw a circle which ſhall touch the two auxiliary ones,
and likewiſe paſs through the point A which is the center of the firſt circle. Let the center of this laſt deſcribed circle be E, and the ſame point E will like-
wiſe be the center of the circle required; as will appear by adding equals to
equals, or taking equals from equals, as the caſe requires.

The Caſes are theſe.

Case 1ſt. Iſ it be required that the circle ſhould touch and include all the other
three; then let A be the center of the greateſt given circle, B of the next, and D
of the leaſt: and let BG = the difference of the ſemidiameters of the 1ſt and 2d,

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