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Case
3d. Suppoſe the given point A to lie in the given circle, whoſe center

is B.

Then
joining AB and continuing it to meet the given circumference in C and

O, biſect AC in E, and from E towards O ſetting off EH = BC the given

Radius, and with A and B Foci and EH tranſverſe Axis deſcribing an Ellipſe

EKH, it will evidently be the Locus required.

## 35. PROBLEM VI.

Having
a right line BC given, and alſo a circle whoſe center is A, to deter-

mine the Locus of the centers of the circles which ſhall be touched both by the

given right line and alſo by the given circle.

There
are three Caſes, but they are all comprchended under one general

ſolution.

Case
1ſt. Let the given right line be without the given circle, and let it be

required that the circles to be deſcribed be touched outwardly by the given circle.

Case
2d. Let the given right line be without the given circle, and let it be

required that the circles to be deſcribed, be touched inwardly by the given

circle.

Case
3d. Let the given right line be within the given circle, and then the

circles to be deſcribed muſt be touched outwardly by the given circle.

## 36. General Solution.

From
the given center A let fall a perpendicular AG to the given line BC,

which meets the given circumference in D [or in Caſes 2d and 3d is produced

to meet it in D] and biſecting DG in F, and ſetting off FM = FA (which is the

ſame thing as making GM = AD the given Radius) and through M drawing

MLK parallel to the given line BC, with A Focus and LK Directrix deſcribe a

Parabola, and it will be the Locus of the centers of the circles required; for

from the property of the Curve FA = FM, and adding equals to equals, or

ſubtracting equals from equals, as the Caſe requires, FD = FG.