Monday, October 22, 2012

3 Points Make a Circle

One of the problems presented in investigation 3 was this: "given three points, [find] the center of a circle that would pass through those three points." This question was posed in four different contexts, including compass and straightedge, ruler and protractor, paper folding and--the one that we chose--an algebraic approach.

So what are the things that we already do know? Well the problem states that we are already given the x and y coordinates of the three points that define the circle.  The definition of a circle is "A curve made up of the of all the points on a two dimensional plane that are a fixed distance away from a central point" according to our glossary of objects. So that means that the length of the lines drawn from each of these points to the unknown center are all the same length.

Point A is unknown. Coordinates of points B, C, and D are all known. All lines between points are equidistant from the center.

So, now we have to figure out the lengths of the lines in terms of the points they are connected to. We can do this by using the pythagorean theorem (I'm just gonna take it for granted this time; feel free to write a blog post proving the theorem if you like!). This example is for any of three points where its 2 coordinates are A and B, the points of the unknown are x and y, and the length of the line is l.

|A-x|^2 + |B-y|^2 = l^2

The absolute value signs are there to account for negative values, because in this case we don't care if line is pointing to the left; we just need to know total heftiness of where it is going (its length): its absolute value. Now all we have to do is plug this in for the other two points. Keep in mind that this does not correspond to the diagram in terms of letters; the capital letters are just the x and y coordinates of the 3 points that we already know. Also remember that throughout these 3 equations x and y remain the same.

|A-x|^2 + |B-y|^2 = l^2
|C-x|^2 + |D-y|^2 = l^2
|E-x|^2 + |F-y|^2 = l^2

This system of equations really shows the relationship between these four points. Now it is purely a matter of only Algebra. But we wanted a little more satisfaction (maybe get rid of the absolute values?), so we plugged this system of equations into Wolfram Mathematica. Well, let me just say that the output was less than satisfactory. There is even a couple of is in there(i is the square root of -1: a scary number). Unfortunately, I don't have the Wolfram Mathematica output with me now, so I will upload it later.

Ending this problem with a slightly bitter taste in my mouth, I wonder if taking a compass and straightedge (or other approach) to this problem will give more elegant results. What if you try the equation with only 2 points? My guess is that the graph of the equations plots a line where the center could be, instead of a single point. But that's just a hunch. It is up to you, dear reader to investigate!

--Solution by Harry LR and Akash V Mehta
--Write up by HLR

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