Circle through three points: centre, radius and equation

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.

Updated 27 September 2026Checked against 4 worked examples

Radius

Shown to up to 8 decimal places, half-up

Centre x: 1
Centre y: 1
Standard form: (x − 1)² + (y − 1)² = 25
General form: x² + y² − 2x − 2y − 23 = 0
Diameter: 10
Circumference: 31.4159
Area: 78.5398

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.

Circle through the three points

How it's calculated S

Determinant (zero means the points are collinear)
$D = 2[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] = -120$
Centre
$\begin{gathered} h = \frac{(x_1^2 + y_1^2)(y_2 - y_3) + (x_2^2 + y_2^2)(y_3 - y_1) + (x_3^2 + y_3^2)(y_1 - y_2)}{D} = 1 \\[6pt] k = \frac{(x_1^2 + y_1^2)(x_3 - x_2) + (x_2^2 + y_2^2)(x_1 - x_3) + (x_3^2 + y_3^2)(x_2 - x_1)}{D} = 1 \end{gathered}$
The centre is where the perpendicular bisectors of P₁P₂ and P₂P₃ meet, so it is equally far from all three points.
Radius
$r = \sqrt{(x_1 - h)^2 + (y_1 - k)^2} = \sqrt{25} = 5$
Equation
Standard: (x − 1)² + (y − 1)² = 25 · General: x² + y² − 2x − 2y − 23 = 0

About circle through three points

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.

Worked examples

(−3, 4), (4, 5), (1, −4)

Point 1: x: -3
Point 1: y: 4
Point 2: x: 4
Point 2: y: 5
Point 3: x: 1
Point 3: y: -4

Centre x: 1
Centre y: 1
Radius: 5
Standard form: (x − 1)² + (y − 1)² = 25
General form: x² + y² − 2x − 2y − 23 = 0

Checked against: Python 3.8 fractions with the determinant formula; each point is 5 from (1, 1): 4² + 3², 3² + 4², 0² + 5²

Right-angled corner (0, 0), (4, 0), (0, 3)

Point 1: x: 0
Point 1: y: 0
Point 2: x: 4
Point 2: y: 0
Point 3: x: 0
Point 3: y: 3

Centre x: 2
Centre y: 1.5
Radius: 2.5
Area: 19.634954

Checked against: Thales: the hypotenuse (length 5) is a diameter, centre at its midpoint; Python 3.8 math: pi*2.5**2

Centre at the origin (edge case: no shift terms)

Point 1: x: 1
Point 1: y: 0
Point 2: x: 0
Point 2: y: 1
Point 3: x: -1
Point 3: y: 0

Centre x: 0
Centre y: 0
Radius: 1
Standard form: x² + y² = 1
General form: x² + y² − 1 = 0

Checked against: All three points are 1 from the origin

Fractional centre (0, 0), (2, 0), (1, 3)

Point 1: x: 0
Point 1: y: 0
Point 2: x: 2
Point 2: y: 0
Point 3: x: 1
Point 3: y: 3

Centre x: 1
Centre y: 1.33333333
Radius: 1.66666667
Standard form: (x − 1)² + (y − 4/3)² = 25/9

Checked against: Python 3.8 fractions: h = 1 by symmetry, 9 − 6k = 1 gives k = 4/3, r² = 1 + 16/9 = 25/9

Questions

How do you find the equation of a circle through three points?

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.

How do you find the centre of a circle from three points on it?

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.

What is the general form of the equation of a circle?

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.

Why is there no circle through three points on a straight line?

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.

How accurate is the circle through three points calculator?

Accuracy depends on your inputs and the method's assumptions. Decimal arithmetic uses 50 significant digits, but estimates, numerical methods and source data can be less precise; the displayed rounding does not remove those limits. It is checked against 4 worked examples whose answers come from independent sources; for example, “(−3, 4), (4, 5), (1, −4)” is checked against Python 3.8 fractions with the determinant formula; each point is 5 from (1, 1): 4² + 3², 3² + 4², 0² + 5².

Where does the method come from?

Weisstein, E. W. “Circumcircle” — MathWorld (circle through three points, determinant form); Weisstein, E. W. “Circle” — MathWorld (standard and general equations).

About this calculator

\begin{gathered} h = \frac{\sum (x_i^2 + y_i^2)(y_j - y_k)}{2\sum x_i(y_j - y_k)},\quad k = \frac{\sum (x_i^2 + y_i^2)(x_k - x_j)}{2\sum x_i(y_j - y_k)} \\[6pt] (x - h)^2 + (y - k)^2 = r^2 \end{gathered}

Sources

Checked against references

4 worked examples with independently sourced answers ship with this calculator. They run in the test suite; you can run them here too.