I have a question regarding the geometric properties of a circle.
For any circle, when taking the radius from the center of the circle, the radius will be constant throughout the whole circle.
However, my given issue is occurs when the radius is not taken at the center of the circle. Assuming that the radius of the circle is offset such that the radius is taken from a point below the circle instead. Assuming that the radius of the circle is now taken at every 45 degree interval, resulting in 8 different radiuses.
Using these 8 given radiuses, would it be possible to deduce the circumference of the circle?
Thanks for reading through, and appreciate any help that can be given.
Best Answer
If the distances from the point within the circle are in fact arranged every 45º around that point, there should be three pairs of measurements with identical values within each pair, owing to the symmetry of the circle. Two measurements will not be paired -- the shortest and longest among them. These would be 180º apart, so they fall on a diameter of the circle: the radius of the circle is half the sum of those lengths. The center is midway between the location of those endpoints. (Not much of this would be true in general if the curve weren't a circle.)
Is this the situation you're asking about?
Or do you mean the 8 points are spaced equally around the circumference?
This is a bit inelegant, but here's something you can try. If you know the relative positions of the eight points around the circle, take any three that are 45º apart and make them "cardinal points" $ \ (R, \ 0) \ , \ ( \frac{\sqrt{2}}{2}R, \ \frac{\sqrt{2}}{2}R) \ $ and $ \ (0, \ R) \ $ . Let the interior point of the circle be $ \ (X,Y) \ $ . If we call the known measurements from $ \ (X, \ Y) \ $ to each of these points $ \ s_1 \ , \ s_2 \ $ and $ \ s_3 \ $ , then we can write
$$ (X \ - \ R)^2 \ + \ Y^2 \ = \ s_1^2 \ \ , \quad \mathbf{[1]} $$
$$ (X \ - \ \frac{\sqrt{2}}{2}R)^2 \ + \ (Y \ - \ \frac{\sqrt{2}}{2}R)^2 \ = \ s_2^2 \ \ , \quad \mathbf{[2]} $$
$$ \ \ X^2 \ + \ (Y \ - \ R)^2 \ = \ s_3^2 \ \ . \quad \mathbf{[3]} $$
We can then solve these equations simultaneously for the radius of the circle and the coordinates of the unspecified point relative to the center of the circle. (I initially tried this using the quadrant points, but too many terms cancel out.)
EDIT (6/19) --
I should mention that it is not easy to simply "make up a problem" for this method. While it is the case that three non-collinear points define a circle, and choosing three such points at random will lead to a solution circle, it is not simple to pick three lengths and expect to obtain a "radiant point" and a radius for the circle in this "inverse problem".
I found that I had to draw a circle and take measurements in order to test this approach (just using randomly-selected length measurements can lead to a non-solution with inconsistent results). For the measurements $ \ s_1 \ = \ 3.5 \ , \ s_2 \ = \ 3.7 \ , \ s_3 \ = \ 7.55 \ , \ $ we can solve the system of equations 1 - 3 above as:
$$ (X \ - \ R)^2 \ + \ Y^2 \ = \ 3.5^2 \ \ \Rightarrow \ \ X^2 \ + \ Y^2 \ + \ R^2 \ = \ 3.5^2 \ + \ 2RX \quad \mathbf{[A]} \ , $$
$$ X^2 \ + \ (Y \ - \ R)^2 \ = \ 7.55^2 \ \ \Rightarrow \ \ X^2 \ + \ Y^2 \ + \ R^2 \ = \ 7.55^2 \ + \ 2RY $$
$$ \Rightarrow \ \ 2RX \ - \ 2RY \ = \ 7.55^2 \ - \ 3.5^2 \ \ \Rightarrow \ \ RX \ - \ RY \ \approx \ 22.376 \ \ ; $$
$$ (X \ - \ \frac{\sqrt{2}}{2}R)^2 \ + \ (Y \ - \ \frac{\sqrt{2}}{2}R)^2 \ = \ 3.7^2 $$
$$ \Rightarrow \ \ X^2 \ + \ Y^2 \ + \ R^2 \ = \ 3.7^2 \ + \ \sqrt{2} \ RX \ + \ \sqrt{2} \ RY $$
$$ \Rightarrow \ \ 3.5^2 \ + \ 2RX \ \approx \ 3.7^2 \ + \ \sqrt{2} \ RX \ + \ \sqrt{2} \ (RX \ - \ 22.376) $$
$$ \Rightarrow \ \ RX \ \approx \ 36.478 \ \ , \ \ RY \ \approx \ 14.102 \ \ . $$
We can now insert these results into, say, equation $ \ \mathbf{A} \ $ to obtain
$$ \left( \frac{36.478}{R} \right)^2 \ + \ \left( \frac{14.102}{R} \right)^2 \ + \ R^2 \ = \ 3.5^2 \ + \ 2 \ \cdot \ 36.478 $$
$$ \Rightarrow \ \ \frac{1529.52}{R^2} \ + \ R^2 \ \approx \ 85.206 \ \ \Rightarrow \ \ R^4 \ - \ 85.206 \ R^2 \ + \ 1529.52 \ = \ 0 \ \ . $$
The polynomial function in this equation has even symmetry, so its four (real) solutions are $ \ \approx \ \pm 5.070 \ , \ \approx \ \pm 7.714 \ $ . Since the values are supposed to represent the radius of a circle, we discard the negative solutions and may now conclude
$$ R \ \approx \ 5.070 \ \ , \ \ (X, \ Y) \ \approx \ ( \ 7.195, \ 2.781 \ ) \ \ , $$
$$ R \ \approx \ 7.714 \ \ , \ \ (X, \ Y) \ \approx \ ( \ 4.729, \ 1.828 \ ) \ \ . $$
There are two solutions: upon graphing them, we see that the smaller circle [shown in red below] provides the result for which the "radiant point" is external to the circle, while the larger circle [in blue] has the internal radiant point.
A striking feature of these paired circles is that the radiant point for the segments extending to the points on the smaller circle appears to lie on the larger circle, and vice versa. Having worked on a couple other examples, this does not seem to be a coincidence. I suspect there is a theorem underlying this, but I haven't taken time to prove it yet (it is likely a known result in circle geometry, though I'm not familiar with it...).
One may well modify the above calculations appropriately to work with other choices among the eight points.
[Now I feel a bit better about receiving the up-votes and "checkmark".]