Bisect DF at A. Also about D describe an arc
with any radius DP greater than DA, and about O another arc with a
radius OP = DP + FO, intersecting the first arc at P, then draw PD,
and also PO, cutting the circumference of the given circle in L. Since
PD = PL, and DA = AF, it is evident that by repeating this process we
shall construct a curve PAR, which satisfies the condition that _every
point in it is equally distant from a given point and from the
circumference of a given circle_. Since PO-PD = LO, and AO-AD = FO,
this curve is one branch of the hyperbola of which D and O are the
foci.
[Illustration: FIG. 1]
Bisect DG at B, then about D describe an arc with any radius DQ
greater than DB, and about O another are with radius OQ = DQ-FO; draw
from Q the intersections of these arcs, the line QD, and also QO,
producing the latter to cut the circumference in E. By this process we
may construct the curve QBZ, each point of which is also equally
distant from the given point D, and from the concave instead of the
convex arc of the given circumference. The difference between QD and
QO being constant and equal to FO, and AB being also equal to FO, this
curve is the other branch of the same hyperbola, whose major axis is
equal to the radius of the given circle.
The tangent at P bisects the angle DPL, and is perpendicular to DL,
which it bisects at a point I on the circumference of the circle whose
diameter is AB, the major axis, the center being C, the middle point
of D O.
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