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

let another circle be deſcribed, touching this laſt, and alſo the line XH and paſſ-
ing through the point A by Problem X, and I ſay that E the center of this circle
will alſo be the center of the circle required; as will appear by taking equals from
equals, or adding equals to equals, as the aſſigned Caſe and Data ſhall require.

The Caſes are four, though Vieta makes but three.

Case Iſt. If it be required that the circle ſhould touch both the others ex-
ternally then BG muſt be taken equal to the difference of the Semidiameters of
the two given circles, and ZX muſt be taken in BZ produced.

Case 2d. If it be required that the circle ſhall touch and include both the
given ones; then BG muſt be taken equal to the difference, as in Caſe 1ſt, but
ZX muſt be taken in BZ itſelf.

Case 3d. If it be required that the circle ſhould touch and include the greater
of the given circles, and touch externally the other whoſe center is B; then BG
muſt be taken equal to the ſum of the Radii of the given circles, and ZX muſt
be taken in BZ itſelf.

Case 4th. If it be required that the circle ſhould touch the greater of the
given circles externally, and touch and include the leſſer; then BG muſt be taken
equal to the ſum of the Radii, and ZX muſt be taken in BZ produced.

## 25.PROBLEM XII .

Having two points given B and D, and like wiſe a circle whoſe center is A; to de-
ſcribe another circle which ſhall paſs through the given points, and touch the
given circle.

Let DB be joined, as alſo AB, and let AB be produced to cut the given
circle in the points I and K, then let BH be taken a 4th proportional to DB,
BK, BI; ſo that BD X BH = BI X BK: from H let a Tangent HF be drawn
to the given circle; and BF be joined and cut the circle again in G: and let DG
be drawn cutting the given circle again in E, and laſtly through the points D,
B, G, let a circle be drawn, I ſay it will touch the given circle in G.

For joining EF; becauſe the rectangle DBH = the Rectangle KBI, i. e. the
Rectangle GBF, thereforethefourpoints D, H, F, G, are in a circle; and hence the
angle HFB = the angle GDB: (for in the two firſt ſigures one is theexternal angle
of a quadrilateral figure, and the other is the internal and oppoſite, and in the two
laſt figures theſe angles are in the ſame ſegment.) But the angle HFB = the angle
GEF by Eu. III. 32. hence GEF = GDB: therefore the triangles GEF and
GDB are ſimilar and under the ſame vertex, and therefore by Lemma 3. the cir-
cles deſcribed about them will touch each other in the common vertex G.

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