**PARABOLA CONSTRUCTION WITH A CIRCLE AS THE
DIRECTRIX**

QUESTION: Construct the locus of points equidistant
from a fixed point F and a circle. In other
words, repeat the parabola construction but
use a circle as the "directrix."
Let F be any point in the plane other than
the center of the circle. Assume F is not
on the circle; it can be either inside or
outside.

STEP ONE: Create one point F (focus) and
one circle such that point F is different
than center O of the circle .

STEP TWO: Create a movable point X on the
circle.

STEP THREE: Construct the line that passes
through points O and X.

STEP FOUR: Construct the line segment [FX].
Locate its midpoint M.

STEP FIVE: Construct the perpendicular bisector
b of the line segment [FX]

STEP SIX: Construct the intersection point
T of the line OX and the bisector b.

STEP SEVEN: Now [TF]=[TX] by construction.
Therefore T will represent our trace point.

CONSTRUCTION IS COMPLETE. LET'S SEE NOW THE
LOCUS OF THE TRACE POINT T:

CASE ONE: F IS OUTSIDE THE CIRCLE: **T TRACES OUT A HYPERBOLA WITH FOCI F AND
O**.

CASE TWO: F IS ON THE CIRCLE: **A TRIVIAL CASE: T COINCIDES WITH O: T TRACES
OUT NOTHING!**

CASE THREE: F IS INSIDE THE CIRCLE AND DOES
NOT COINCIDE WITH O: **T TRACES OUT AN ELLIPSE WITH FOCI F AND O.**

CASE FOUR: F IS INSIDE THE CIRCLE AND DOES
COINCIDE WITH O: **T TRACES OUT A CIRCLE CENTERED AT O WITH
RADIUS HALF THE RADIUS OF THE ORIGINAL CIRCLE.**

In this way, I explored all the possible
cases (including the trivial ones). Click
here to download the GSP file.