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.