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### Analytic Proof that Midpoints Form a Circle

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Date: 03/10/98 at 06:10:03
From: Allan Dunlavy
Subject: Circle Geometry

I am doing a math project that investigates a "What if....?" game.
This game takes a circle and then you choose a point P within the
circle. Lines are then drawn from this point to points on the edge of
the circle. I found that the midpoints of these lines also formed a
circle.

I am trying to find out why this occurs and how I prove this
mathematically. So far, I have been investigating the idea of proving
it using coordinate geometry. Unfortunately, I seem to have reached a

Yours sincerely,

Allan
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Date: 03/10/98 at 12:41:36
From: Doctor Rob
Subject: Re: Circle Geometry (Allan Dunlavy)

Allan,

You are correct that you get another circle; and, in fact, the radius
of the new circle is exactly half of the radius of the old circle,
and the center of the new circle is halfway between the point P and
the center of the old circle. Moreover, the point doesn't have to be
inside the old circle!

You can prove this using analytic geometry as follows.

Let the point be P(0,0), and the circle be given by the pairs (x,y)
that satisfy the equation

(x-h)^2 + (y-k)^2 = r^2,

with radius r and center (h,k).

Now the point midway between P and (x,y) has coordinates (x/2,y/2).
Call these x' = x/2 and y' = y/2. Divide the equation of the old
circle by 4, and replace x with 2*x' and y with 2*y'. This gives you
the equation

(x'-h/2)^2 + (y'-k/2)^2 = (r/2)^2,

which the pair (x',y') satisfies. This is the equation of a circle
with center (h/2, k/2) and radius r/2. Since this works for any (x,y)
on the old circle, the new circle is the set of midpoints you sought.

By the way, P is inside the circle if and only if h^2 + k^2 < r^2.
Since this condition was not stipulated above, it is immaterial
whether P is inside the circle or not.

Another proof can be obtained by considering the old circle to be the
base of an oblique circular cone, and the point P to be the vertical
projection of the vertex of the cone onto the plane of the base.

Then the new circle is the projection of the intersection of the cone
with a plane parallel to the base and halfway between the base and the
vertex, which is obviously a circle of half the radius of the base.

Since the projection of a circle is a circle of the same radius, you
have your result. Your line segments are then the projections of the
line segments from the vertex of the cone to points on the circle
forming the base.

-Doctor Rob, The Math Forum
Check out our web site http://mathforum.org/dr.math/
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Associated Topics:
College Conic Sections/Circles
High School Conic Sections/Circles

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