Case
5th. Let one of the given circles include the other, and let be required

that the circle to be deſcribed be touched outwardly by one of the given circles,

and inwardly by the other.

Limitation
. Then it’s Diameter muſt not be given greater than the greater

ſegment of the right line, joining the centers of the given circles, which is in-

tercepted between the two concave circumferences of the ſaid circles, nor leſs

than the leſſer ſegment; viz. not greater than CD, nor leſs than MN.

Case
6th. Let the two given circles cut each other, and let it be required

that the circle to be deſcribed, and to be touched by them both, ſhall alſo be

included in each of them.

Limitation
. Then it’s Diameter muſt not be given greater than the ſeg-

ment of the right line, joining the centers of the given circles, intercepted by

their concave circumferences, which lies in the ſpace common to both the given

circles; viz. not greater than CD.

There
may be alſo three other Caſes of this Problem, when the given circles

cut each other; but becauſe they are ſimilar to the 1ſt, 2d, and 4th Caſes already

propoſed, and ſubject to juſt the ſame Limitations; except that which is ſimilar

to the 1ſt, which is ſubject to no Limitation at all, they are here omitted; as are

likewiſe thoſe Caſes where the given circles touch each other; becauſe they are

the ſame as the preceding, and ſolved in the ſame manner.

##
10.
The
GENERAL
Solution
.

Join the given centers A and B, and where the Caſe requires, let AB be pro-

duced to meet the given circumferences in C and D: and let CI and DH be

taken equal to the given line Z: and let two circles be deſcribed; one with cen-

ter A and diſtance AI, and the other with center B and diſtance BH: and theſe

two circles will neceſſarily cut or touch each other by the Limitations given. Let

the point of concourſe be E: from E draw the right line EAF cutting the circle

whoſe center is A in F; as alſo EBG cutting the circle whoſe center is B in G: then with center E and diſtance EF deſcribe a circle FK, this will be the circle

required: becauſe AF and AC are equal as alſo AI and AE; therefore FE

and CI are alſo equal: but CI was made equal to Z, therefore FE is equal to

Z. Again, becauſe BD and BG are equal, as alſo BH and BE, therefore DH

and EG are alſo equal: but DH was made equal to Z, therefore EG is equal to

Z. Hence it appears that the circle FK, paſſing through F will alſo paſs thro’

G, and likewiſe that it will alſo touch the given circles in F and G, becauſe EAF

and EBG are right lines paſſing through the centers.

##
11.
PROBLEM IV.

Having
a given point A, and a given right line BC, it is required to draw a

circle, whoſe Radius ſhall be equal to a given line Z, which ſhall paſs through

the given point, and alſo touch the given line.