If two circles are concentric, then the sum of the squares of the distances from any point of one of them to the endpoints of any diameter of the other, is a fixed quantity.

I'm having a really hard time with this one. For starters, I know that there are two separate cases (where the diameter is in the inner circle, and then when the diameter is in the outer circle). I also know that I can use the theorem "The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides", but I can't seem to figure out exactly how to create a parallelogram from the various situations.


1 Answer 1


It's not so difficult if you make a good drawing. First draw a picture with two concentric circles in the xy plane. For easyness sake, draw one circle with radius $1$ and the other one radius $3$. In the small circle, draw a diameter that makes a 45 degree angle (with positive x-axis). Let's call the end points $A$ and $B$ Second, draw any point $C$ in the 4th quadrant on the bigger circle AND draw a point $D$ in the 2nd quadrant also on the bigger circle exactly opposite of the point in the 4th quadrant. Now I want you to connect these four points to form a quadrilateral. Note that $AB$ is the diameter of the smaller circle and $CD$ the diamter of the bigger circle. The diameters bisect each other. Take a close look at the quadrilateral $ACBD$ What do you notice? How does that property of the parallelogram's sides and diagonals help you here? The other case goes similar.

  • $\begingroup$ Did you mean diameter instead of radius, i.e. AB is the diameter of the smaller circle and CD is the diameter of the larger circle? $\endgroup$ Oct 25, 2015 at 15:18
  • $\begingroup$ @analysischallenged. Yes, you are right, double checked my picture too:) . I changed it. Do you understand my approach now? Why is the quadrilateral a parallelogram? $\endgroup$
    – imranfat
    Oct 25, 2015 at 15:20
  • $\begingroup$ If a quadrilateral has diagonals which bisect each other, then the quadrilateral is a parallelogram. This was a theorem that I proved on a previous assignment. My pictures never looked right because I wasn't drawing a diameter in the larger circle, so I only had a triangle. And then I would select a random second point on the larger circle instead of, as in your example, the specific point D. It all makes sense now. Thanks!! $\endgroup$ Oct 25, 2015 at 15:33
  • $\begingroup$ ok, you're welcome (good post by the way) $\endgroup$
    – imranfat
    Oct 25, 2015 at 15:40

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