Centre, radius and equation of the circle through three points, in standard form (x − h)² + (y − k)² = r² and general form, plus area.
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The only circle through (−3, 4), (4, 5) and (1, −4) is centred at (1, 1) with radius 5: (x − 1)² + (y − 1)² = 25.
The centre is where the perpendicular bisectors of P₁P₂ and P₂P₃ meet, so it is equally far from all three points.
Standard: (x − 1)² + (y − 1)² = 25 · General: x² + y² − 2x − 2y − 23 = 0
Three points that do not lie on one line fix exactly one circle, the circumcircle of the triangle they form. Its centre (h, k) is where the perpendicular bisectors of two chords meet, which a determinant formula gives directly, and the radius is the distance from the centre to any of the points. Coordinates stay exact fractions, so a centre at (1, 4/3) is shown as 4/3, not 1.3333.
The default points (−3, 4), (4, 5) and (1, −4) give centre (1, 1) and radius 5: (x − 1)² + (y − 1)² = 25, or x² + y² − 2x − 2y − 23 = 0. The same construction finds the centre of a round table, pipe or arch from three marks on its edge, and the radius of a road curve from three survey points.
If the determinant is zero the points are collinear and no circle exists. Coordinates carry no unit; the radius and area are in the coordinates' unit and its square.
Three points not on one line lie on exactly one circle, the circumcircle of their triangle. Its center is where the perpendicular bisectors of two sides cross, and the radius reaches each point: (0, 0), (2, 0) and (1, 3) give center (1, 4/3) and radius 5/3.
Each perpendicular bisector is the set of points equally far from both ends of a side, so the point where two of them cross is equally far from all three corners. For (0, 0), (2, 0) and (1, 3):
Solving the bisector equations once in general gives the center (h, k) directly. Let d = 2[x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)]. Then:
The bracket in d is twice the triangle's signed area, so d is zero exactly when the points are collinear. For the example, d = 12, h = 12/12 = 1 and k = 16/12 = 4/3.
With side lengths a, b and c and triangle area K, the circumradius is R = abc/(4K), as given in MathWorld's circumcircle entry. The example triangle has sides 2, √10 and √10 and area 3, so R = 20/12 = 5/3.
The standard form is (x − h)² + (y − k)² = r², and its right side is the radius squared. The example circle is (x − 1)² + (y − 4/3)² = 25/9.
Expanding gives the general form x² + y² + Dx + Ey + F = 0, with D = −2h, E = −2k and F = h² + k² − r². Here F = 0 because the circle passes through the origin: x² + y² − 2x − (8/3)y = 0.
| Three points | Center | Radius | Standard equation |
|---|---|---|---|
| (−3, 4), (4, 5), (1, −4) | (1, 1) | 5 | (x − 1)² + (y − 1)² = 25 |
| (0, 0), (4, 0), (0, 3) | (2, 1.5) | 2.5 | (x − 2)² + (y − 1.5)² = 6.25 |
| (1, 0), (0, 1), (−1, 0) | (0, 0) | 1 | x² + y² = 1 |
| (0, 0), (2, 0), (1, 3) | (1, 4/3) | 5/3 | (x − 1)² + (y − 4/3)² = 25/9 |
To check the first row, (−3, 4) sits 4 left of and 3 above (1, 1), and 4² + 3² = 25; the other two points also give 25. In general form that circle is x² + y² − 2x − 2y − 23 = 0.
The second row has a right angle at (0, 0), so the hypotenuse from (4, 0) to (0, 3) is a diameter and the center is its midpoint.
The center lies inside an acute triangle, at the midpoint of the hypotenuse of a right triangle, and outside an obtuse one.
The points (0, 0), (6, 0) and (1, 1) have an obtuse angle at (1, 1), and their center, (3, −2), lies below all three, with radius √13 ≈ 3.605551.
Use one flat coordinate system and one length unit for all six numbers; the radius comes out in that unit and the circle's area in its square. Latitude and longitude need projecting onto a flat grid first.
For a circle known by its radius, diameter or circumference, use the circle calculator. For distances and midpoints between two points, use the coordinate geometry calculator.
Checked against: With the determinant formula; each point is 5 from (1, 1): 4² + 3², 3² + 4², 0² + 5²
Checked against: Thales: the hypotenuse (length 5) is a diameter, centre at its midpoint
Checked against: All three points are 1 from the origin
Checked against: h = 1 by symmetry, 9 − 6k = 1 gives k = 4/3, r² = 1 + 16/9 = 25/9
Substitute each point into the general form x² + y² + Dx + Ey + F = 0 and solve the three linear equations for D, E and F. For (−3, 4), (4, 5) and (1, −4) this gives D = −2, E = −2 and F = −23. Completing the square turns that into (x − 1)² + (y − 1)² = 25: centre (1, 1), radius 5.
Construct the perpendicular bisectors of two chords, such as P₁P₂ and P₂P₃; they cross at the centre, because every point on a perpendicular bisector is equally far from both ends of its chord. For a right triangle the centre is the midpoint of the hypotenuse: (0, 0), (4, 0) and (0, 3) give centre (2, 1.5) and radius 2.5.
x² + y² + Dx + Ey + F = 0. The centre is (−D/2, −E/2) and the radius is √(D²/4 + E²/4 − F). For x² + y² − 2x − 2y − 23 = 0 the centre is (1, 1) and the radius √(1 + 1 + 23) = 5. If D²/4 + E²/4 − F is zero the equation describes a single point, and if it is negative, no real circle.
A circle meets a straight line at most twice, so it cannot pass through three collinear points. In the formula, x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂), which is twice the triangle's signed area, becomes zero and the centre would need a division by zero. Points that are nearly collinear give a very large radius.
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