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Proof that 3 non-collinear points determine a unique circle

Proof of the Assertion that Any Three Non-Collinear Points Determine Exactly One Circle
This is an interesting problem in geometry, for a couple of reasons. First, you can apply some earlier, basic geometry principles; and secondly, you can choose two different strategies for solving the problem.

The basic geometry underlying: any three non-collinear points determine a plane, somewhere in 3D space. Once that has been done, imagine that the plane has been rotated into the x-y plane, which will make the problem much easier to solve!

The two strategies for solution are: (Proof A) actually solve to find the circle. This is equivalent to finding the center of the circle (finding the equation of the circle is simple from there). But, you actually have to do some math to get this! If, while doing this, there is no possibility to obtain other values for the coordinates of the center of the circle, you have proved the assertion as well as obtained a method (and perhaps the formula) for doing so.
The other possibility (Proof B) is to assume that one center of such a circle exists, then show that another center of such a circle (i.e. the center of a different circle) could not exist for some reason. This is mathematically fun to do!

So, (Proof B) runs as follows: let the three original non-collinear points be called A, B, and C. Let the proposed centers of the two different circles be designated as R1 and R2 (note that these are not radii, but actual points). Then, from the basic properties of a circle, any radius has the same length. Expressing each of the radii AR1, BR1, and CR1 by the Pythagorean theorem, using x(A) to be the x-coordinate of point A, and y(A) to be the y-coordinate of point A, etc.:

((x(A) – x(R1))^2 + ( (y(A) – y(R1))^2 =
((x(B) – x(R1))^2 + ( (y(B) – y(R1))^2 =
((x(C) – x(R1))^2 + ( (y(C) – y(R1))^2

and also:

((x(A) – x(R2))^2 + ( (y(A) – y(R2))^2 =
((x(B) – x(R2))^2 + ( (y(B) – y(R2))^2 =
((x(C) – x(R2))^2 + ( (y(C) – y(R2))^2

Now, these two sets of equalities look like three equations in 2 unknowns – i.e. overspecified! But this is not the case, because the equations are not equal to a constant, but to each other. So, each pair of two equations has infinitely many solutions – the line bisecting the segment joining the respective specified points – and the third equality serves to uniquely “fix” one point as the common intersection of the three such lines generated from each pair of points in the set {A, B, C} (there, now, you have the essential idea for calculating the center. But, leave that alone for now.).

Now, these look like two formidable sets of equations to start drawing conclusions from! But, they are related, because the two proposed circle center points have an offset from each other, in possibly both their x and y coordinates. Therefore,
Let x(D) = x(R1) – x(R2) and
y(D) = y(R1) – y(R2)

Then substitute these terms into the *second* set of three equations, to express them ONLY in terms of R1’s coordinates:

(x(A) – x(R1) + x(D))^2 + ((y(A) – y(R1) + y(D))^2 =
(x(B) – x(R1) + x(D))^2 + ((y(B) – y(R1) + y(D))^2 =
(x(C) – x(R1) + x(D))^2 + ((y(C) – y(R1) + y(D))^2

Now, these equations are somewhat similar to the first set. Capitalize on the similarity, by considering each *(P) - *(R1) expression as a single term, but otherwise expanding the squares as binomial expansions (you’ll see why, in a minute):

(x(A) – x(R1))^2 + 2x(D)(x(A) – x(R1)) + (x(D))^2 + (y(A) – y(R1))^2 + 2y(D)(y(A) – y(R1)) + (y(D))^2 =
(x(B) – x(R1))^2 + 2x(D)(x(B) – x(R1)) + (x(D))^2 + (y(B) – y(R1))^2 + 2y(D)(y(B) – y(R1)) + (y(D))^2 =
(x(C) – x(R1))^2 + 2x(D)(x(C) – x(R1)) + (x(D))^2 + (y(C) – y(R1))^2 + 2y(D)(y(C) – y(R1)) + (y(D))^2


Now, compare the three equations you just derived. In common, they each have a term:
((x(P) – x(R1))^2 + ((y(P) – y(R1))^2 (where P is one of the three original points)
and also a term:
(x(D))^2 + ((y(D))^2

We can drop (subtract out) both of these types of terms from the equalities, because, by our original conditions, the first set of terms just restates that the three points are all equidistant from point P1 (by the Pythagorean theorem), and the second set of terms are all a common constant (the square of the distance between the two proposed circle centers).

Then, we are left with only the middle terms of the binomial expansions:

2x(D)(x(A) – x(R1)) + 2y(D)(y(A) – y(R1)) =
2x(D)(x(B) – x(R1)) + 2y(D)(y(B) – y(R1)) =
2x(D)(x(C) – x(R1)) + 2y(D)(y(C) – y(R1))

These are really simplifying nicely, but we’re not done yet; take out the remaining common terms, and divide through by 2, to obtain:

x(D)x(A) + y(D)y(A) = x(D)x(B) + y(D)y(B) = x(D)x(C) + y(D)y(C)

Do you realize what you have here? It’s a set of equations, where x(D) and y(D) are the constants “a” and “b” (assuming x(D) and y(D) are non-zero values), all saying ax + by = some common value -- this is the equation for a single line. More importantly, since coordinates for points A, B, and C all satisfy the equation, all three points must lie on that line! But, we stated above, A, B, and C are *non-*collinear! Therefore, the only way that the equation can be true is if x(D) and y(D) are both zero, meaning that the “second” circle center *must* be the same as the “first” circle center, -- in other words, there can only be one, unique, circle that contains the 3 non-collinear points A, B, and C.

It’s perhaps instructive to consider what would be the meaning of three *collinear* points on a circle. In this case, the circle would have to be of infinite radius, the center to one side or the other of the collinearity line – and in this case, the center could be offset “laterally” (i.e. in a direction parallel to the collinearity line) to any amount one desires, just as the equation above implies.

Comments

You can prove existence and uniqueness of a circle defined by three non-collinear points without using coordinates.
A circle is defined as the set of points that are equidistant from a fixed point, P. Take two points A and B; then the set of points equidistant from both is a line passing through the midpoint of the segment AB and perpendicular to it. Now take a third point C, not collinear with A and B, and construct the line of points equidistant to A and C. Since A, B, and C are not collinear, the two lines are not parallel, so they will intersect in a unique point, P. This point is the unique point which is equidistant from A, B, and C, so it defines the unique circle containing A, B, C.
Well, yes, Andre, constructing the center of the circle implies its uniqueness. And perhaps, the indeterminacy of the "circle center" for three collinear points would be implied thereby. But it's nifty to show it by the alternative route above, i.e. algebraically rather that geometrically!

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