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Special Case: the Circle

In Book IV of his Elements, Euclid gives one way to generalize the result that two distinct points determine a line. Proposition 5 of that book gives the construction of the circle that circumscribes any given triangle. Because a triangle is determined by three noncollinear points, Euclid's proof essentially says:

Theorem 1   Three noncollinear points in the plane determine a unique circle.

Euclid's proof is entirely geometric. Given a triangle \( ABC \), he constructs perpendicular bisectors on two of the sides. The point \( F \) where these two lines intersect is called the circumcenter of \( ABC \) and does not depend on which two sides are chosen. Euclid then shows that the distance from \( F \) to each of the points \( A \), \( B \) and \( C \) is the same, say \( r \). Therefore, the circle with center \( F \) and radius \( r \) passes through all three points \( A \), \( B \) and \( C \).

Figure 2: A triangle and its circumcircle
Open a dynamic GeoGebra applet in a new window]

It's interesting to observe that in the case of an obtuse angled triangle, the circumcenter falls outside of the triangle \( ABC \). Furthermore, if \( ABC \) is a right triangle, then the circumcenter is the midpoint of the hypotenuse, so that the hypotenuse is the diameter of the circle passing through the three points. All of these features can be explored by moving any of the points \( A \), \( B \) or \( C \) in the applet in Figure 2.

After the invention of analytic geometry, it became possible to solve this problem algebraically. A circle with center \( (h,k) \) and radius \( r \) satisfies the equation \[ (x-h)^2 + (y-k)^2 = r^2, \] so given three noncollinear points with coordinates \( (x_1,y_1) \), \( (x_2,y_2) \) and \( (x_3,y_3) \), we substitute these pairs of numbers to get three equations in the three unknowns \( h \), \( k \) and \( r \). Although these equations aren't linear, it's possible to do a little algebra to eliminate the \( r^2 \) and get two linear equations in the unknowns \( h \) and \( k \), which will always have a solution, as long as the given points are not collinear. Rather that work through these details, we'll consider the example of curve fitting with three points and a parabola.