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.