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.