Limitation
. 2Z muſt not be given leſs than the perpendicular let fall from

the given point A upon the given line BC.

From
the point A let AD be drawn perpendicular to BC, and in this perpen-

dicular take DE equal to the given line Z: and through E draw EF parallel to

BC, and from A upon this line EF ſet off AF equal to Z, which may be done,

for by the Limitation Z is not leſs than AE: then with center F and diſtance

FA deſcribe a circle, and I ſay it will touch the line BC: for through F drawing

FG parallel to AD, FGDE will be a Parallelogram, and FG will be equal to

DE, that is to Z, and at right angles to BC.

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12.
PROBLEM V.

Having
a given point A, and alſo a given circle whoſe center is B, it is re-

quired 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 circle.

This
Problem has three Caſes, each of which is ſubject to a Limitation.

Case
Iſt. Let the circle to be doſcribed be required to be touched outwardly

by the given circle.

Limitation
. Then the Diameter muſt not be given leſs than the ſegment

of the right line, joining the given point and the center of the given circle,

which is intercepted between the given point and the convex circumference; viz. not leſs than AC.

Case
2d. Let the circle to be deſcribed be required to be touched inwardly

by the given circle.

Limitation
. Then the Diameter muſt not be given leſs than the right line

which, drawn from the given point through the center of the given circle, is con-

tained between the given point and the concave circumſerence; viz. not leſs

than AC.

Case
3d. Let the given point lie in the given circle.

Limitation
. Then a diameter of the given circle being drawn through the

given point, it is divided into two ſegments by the ſaid point, and the Diameter

of the circle required muſt not be given greater than the greater of them, nor

leſs than the leſſer; viz. not greater than AC, nor leſs than AG.

##
13.
The general
Solution
.

Let
A and B be joined, and in the line AB take CF equal to Z, and then

with center A and diſtance Z, let an arc be drawn, and with center B, and

diſtance BF let another be drawn, which by the Limitations will neceſſarily

either touch or cut the former; let the point of their concourſe be D; then with

D center and DA diſtance let a circle be drawn, and I ſay it will touch the

given circle whoſe center is B: for DB being drawn meeting the circumference

of the circle whoſe center is B in E, BC is equal to BE, and hence CF equals

ED, and they are both equal to the given line Z.