a circle which ſhall paſs through the given point, and likewiſe touch both the

given line and the given circle.

The
given right line may either 1ſt cut, touch, or be entirely without the

given circle, and the given point be without the ſame or in the circum-

ference; or 2dly, it may cut the given circle, and the given point be within

the ſame, or in the circumference.

Case
Iſt. Suppoſe the given line BC to cut, touch, or fall entirely with-

out the given circle; and the given point A to be without the ſame, or in the

circumference: through G the center of the given circle draw DGFC per-

pendicular to BC, and joining DA, take DH a 4th proportional to DA, DC,

DF, ſo that DA X DH = DC X DF: then through the points A and H draw

a circle touching the line CB by Problem VII, and I ſay it will alſo touch the

given circle.

Draw DB cutting the given circle in E, and join FE. Now becauſe the

triangles DEF, DCB are ſimilar, DF: DE: : DB: DC, and therefore DC

X DF = DB X DE. But DC X DF = DA X DH by Conſtruction. Hence

DB X DE = DA X DH, and therefore the points B, E, H, A, will be alſo in

a circle: but the point E is alſo in the given circle; therefore theſe circles either

touch or cut one another in that point. Let now BI be drawn from the point

of contact B perpendicular to the touching line BC to meet the circumference

again in I, and it will be a Diameter: and let EI be joined: then becauſe the

angles FED and BEI are vertical and each of them right ones, FEI will be a

continued ſtraight line: and it appears that the two circles will touch each

other by the preceding Lemmas.

Case
2d. Suppoſe the given line BC to cut the given circle, and the given

point to be within the ſame, or in the circumference; the Conſtruction and De-

monſtration are exactly the ſame as before, except that the angles FED and

BEI are not vertical but coincident, and ſo EI is coincident with EF.

N. B. In either of theſe caſes if the point A coincide with E or be given in

the circumference, draw DEB, and erect BI perpendicular to CB to meet FE in

I, then upon BI as diameter deſcribe a circle, and the thing will manifeſtly be done.

##
24.
PROBLEM XI.

Having
two circles given in magnitude and poſition, whoſe centers are A

and B, as likewiſe a right line CZ; to draw a circle which ſhall touch all three.

From
the center of the leſſer circle B let BZ be drawn perpendicular to CZ,

and in BZ (or in BZ continued as the caſe requires) let be taken ZX = AL the

Radius of the other circle; and through X let XH be drawn parallel to CZ,

and with center B and Radius BG, equal to the difference ( or ſum as the Caſe

requires) of the Radii of the two given circles, let a circle be deſcribed; and laſtly