# Proof of three points are enough to draw one and only one circle

Using the circle theorems or otherwise, I explained why the process locates the centre of the arc. However, I do not know what 'accuracy limitations of this technique' means. I don't think there is any error while explaining the process. Please Help! I used congruence of triangles by drawing perpendicular bisectors to the chord AB and BC.

Process: 1. Draw the line segment AB 2.Locate the label the point M, the midpoint of AB. 3. Draw the line perpendicular to AB through M. 4. Repeat the steps 1 to 3 for points B and C 5. Produce both perpendiculars so that they intersect at O. 6. Verify that the point of intersection, O is the centre of the circle/arc by using OA as the radius and completing the full circumference of the circle.)

Questions: Using the circle theorems of otherwise, explain why this procss locates the centre of the arc.

Discuss the accuracy limitations of this technique.

Thanks!

• "accuracy limitations of this technique" - where did you see this? Anyway, if you're allowed the use of coordinates, you find that the equation of a circle is determined exactly by three conditions, so three points are enough to determine a circle uniquely. – J. M. isn't a mathematician Jun 1 '13 at 4:12
• To add to what J.M. said, those three conditions are: radius, the y-coordinate, and the x-coordinate of the center. – Ataraxia Jun 1 '13 at 4:15
• It is from my folio assignment. I do not know how to show this in coordinates... Do I need to assume the point a(a1,a2) and so on? – pete-mathboy Jun 1 '13 at 4:17
• Could you please give the context, or quote the whole question precisely. – Mark Bennet Jun 1 '13 at 4:19
• sure it is editted – pete-mathboy Jun 1 '13 at 4:25

Limitations may refer to the need for the points to be distinct and non-collinear. Even if a straight line is regarded as a degenerate circle, it does not have a centre or radius. So any formula you find will fail in this case.

• Thank you very much for your help! – pete-mathboy Jun 1 '13 at 4:45

In geometry, there are no accuracy limitations, as our constructions are assumed to be perfect. If there is some error in measuring the location of the points, that can move the center by a large amount. The extreme is three collinear points. If you think of them as on a vertical line, if the middle one moves a bit the center can go from very far away to the right to very far away to the left.
If the lines you use to locate the center are almost parallel, a small error in the point location makes a large error in the center location.

• Thank you very much for your help! – pete-mathboy Jun 1 '13 at 4:45

Regarding "accuracy limitations".

I suppose this refers to things that might go wrong when you move from the ideal perfect world of mathematics to the real world of measurements and computations.

Suppose the three points are very close together, or are very close to being collinear. Then, if there is a tiny error in the location of one of these points, there might be a very large error in the coordinates of the center. Basically, the center location is highly sensitive to the locations of the three points. A numerical analyst would say that the problem is "ill conditioned". These troubles will arise when you are doing floating point calculations on a computer, or when doing constructions with a real compass and ruler.

I used words like "tiny" and "large", which might be enough to answer your question. If you want something more precise, take your formula for the center of the circle, and calculate it's partial derivatives with respect to the coordinates of the input points. You will find that this partial derivative can get very large in some cases. That's the source of the problem.

Or, if you haven't learned about partial derivatives yet, change one of the points by a small amount and recompute the circle center; see how far the center moves.

If we have a (finite) line segment with end points $\,A\,,\,B\,$ , then the perpendicular bisector (p.b.) of the segment $\,AB\,$ is precisely the locus of all points in the plane which are equidistant from points $\,A\,,\,B\,$ , so: when you draw the p.b. of $\,AB\,$ all the points of it are equidistance from $\,A\,$ and from $\,B\,$ , and the same can be said about the p.b. of $\,BC\,$ , so if $\,O\,$ is the intersection point of these p.b.'s, then it is at the same distance from $\,A\,,\,B\;\text{and}\;C\;$ ...