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In Book IV of his *Elements*, Euclid gave 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 constructed 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 showed 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 \(ABC,\) together with its circumcircle \(ABC\) and circumcenter \(F.\) Move points \(A, B\) and \(C\) to explore relationships among triangle, circumcircle, and circumcenter. (Interactive applet created using *GeoGebra.*)

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 than work through these details, we'll consider the example of curve fitting with three points and a parabola.

Robert E. Bradley (Adelphi University) and Lee Stemkoski (Adelphi University), "When Nine Points Are Worth But Eight: Euler’s Resolution of Cramer’s Paradox - Special Case: the Circle," *Loci* (February 2014)