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14.
PROBLEM VI.

Having
a right line given BC, and alſo a circle whoſe center is A, it is re-

quired to draw another circle, whoſe Radius ſhall be equal to a given right line

Z, and which ſhall touch both the given line and alſo the given circle.

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

Case
Iſt, Let the circle to be deſcribed be required to be touched outwardly

by the given circle.

Limitation
. Then the Diameter of the circle required muſt not be given

leſs than the ſegment of a line, drawn from the center of the given circle, per-

pendicular to the given line, which is intercepted between the ſaid line and the

convex circumference; viz. not leſs than BD.

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

the given circle.

Limitation
. Then the given line muſt not be in the given circle, neither

muſt the Diameter of the circle required be given leſs than that portion of the

perpendicular, drawn from the center of the given circle to the given line, which

is intercepted between the ſaid line and the concave circumference; viz. not leſs

than BD.

Case
3d. Let the circle to be deſcribed be required to be both touched and

included in the given circle.

Limitation
. Then the given right line muſt be in the given circle, and

when a Diameter of this given circle is drawn cutting the given line at right an-

gles, the Diameter of the circle required muſt not be given greater than the

greater ſegment ; viz. not greater than BD.

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15.
The general
Solution
.

From
A draw AB perpendicular to BC, cutting the given circumference in D; and in this perpendicular let BG and DF be taken each equal to the given line

Z; and through G draw GE parallel to BC; and with center A and diſtance

AF let an arc be ſtruck, which by the Limitations will neceſſarily either touch

or cut GE; let the point of concourſe be E, let AE be joined, and, if neceſſary,

be produced to meet the given circumference in H; then with E center and

EH diſtance deſcribe a circle, and I ſay it will be the required circle; it is evi-

dent it will touch the given circle: and becauſe AD and AH are equal, as alſo

AF and AE, therefore DF (which was made equal to Z) will be equal to HE: let now EC be drawn perpendicular to BC, then GBCE will be a Parallelogram,

and EC will be equal to GB, which was alſo made equal to Z: hence the

circle will alſo touch the given line BC, becauſe the angle ECB is a right

one.