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,