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Analytic Proof that Midpoints Form a Circle

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 

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 
dead end.

Yours sincerely,    


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


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
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Associated Topics:
College Conic Sections/Circles
High School Conic Sections/Circles

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