Analytic Proof that Midpoints Form a CircleDate: 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 dead end. Yours sincerely, Allan 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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