Dayaan M. answered • 9d
Bachelors in Computer Science with 5 years of tutoring experience
Good question, and the reason every source you found switches to numbers is that the general formula looks unpleasant until you see where it comes from. The derivation is only a few lines, so let me do that first and then the formula will actually mean something.
The idea. The point you want is the circumcenter, and the only thing that defines it is that it is the same distance from all three points. So call it P = (x, y) and set the squared distances equal. Use squared distances rather than distances, because the square roots would make this messy for no reason.
Start with |PA|2 = |PB|2:
(x − x1)2 + (y − y1)2 = (x − x2)2 + (y − y2)2
Expand both sides and watch what happens. The x2 and y2 terms show up on both sides and cancel. That cancellation is the whole trick, and it is why this problem is friendlier than it looks. What survives is linear:
2(x2 − x1)x + 2(y2 − y1)y = (x22 + y22) − (x12 + y12)
That equation is exactly the perpendicular bisector of AB. Doing the same thing with A and C gives
2(x3 − x1)x + 2(y3 − y1)y = (x32 + y32) − (x12 + y12)
So a circumcenter is nothing more than two linear equations in two unknowns. Solve them with Cramer's rule and you get the arbitrary expression you were asking for.
The formula. Write si = xi2 + yi2 to keep things readable, and set
d = 2[x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2)]
Then the center is
x = [s1(y2 − y3) + s2(y3 − y1) + s3(y1 − y2)] / d
y = [s1(x3 − x2) + s2(x1 − x3) + s3(x2 − x1)] / d
Pay attention to the second line, because the x differences run backwards there. It is x3 − x2 and not x2 − x3. That sign flip is the most common place people go wrong when they copy this formula down.
Where the noncollinear condition comes in. That d sitting in the denominator is twice the signed area expression for triangle ABC, so d = 0 happens exactly when A, B and C lie on a single line. The algebra is telling you the same thing the geometry does, which is that three points on a line have parallel perpendicular bisectors and no common circle. Your noncollinear assumption is precisely the condition d ≠ 0.
The radius, if you need it. Once you have the center, r is just the distance out to any one of the three points, so r = √[(x − x1)2 + (y − y1)2]. It does not matter which point you pick, and checking that all three give the same number is a good way to catch an arithmetic slip.
A quick check with numbers. Take A(0, 0), B(4, 0), C(0, 6). Then s1 = 0, s2 = 16, s3 = 36, and d = 2[0(0 − 6) + 4(6 − 0) + 0(0 − 0)] = 48. So x = [0 + 16(6) + 36(0)]/48 = 96/48 = 2, and y = [0(0 − 4) + 16(0 − 0) + 36(4 − 0)]/48 = 144/48 = 3. The center is (2, 3), and the distance from there to each of the three points is √13. That is the midpoint of the hypotenuse, which is exactly where the circumcenter of a right triangle has to be, so the formula holds up.
