# Cercle par trois points : centre, rayon et équation

> Trouvez le centre, le rayon, l'aire et l'équation du cercle passant par trois points, sous la forme (x − h)² + (y − k)² = r² et la forme générale.

Version interactive : https://www.calcopenly.com/fr/geometry/circle-through-three-points
Sujet : Calculatrices de géométrie

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.

## Données

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

## Résultats

- Rayon — résultat principal
- Centre x
- Centre y
- Standard form
- General form
- Diamètre
- Circonférence
- Aire

## Formule

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

## Exemples détaillés

### (−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**
- **Rayon: 5**
- **Standard form: (x − 1)² + (y − 1)² = 25**
- **General form: x² + y² − 2x − 2y − 23 = 0**
- Source de vérification : 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**
- **Rayon: 2.5**
- **Aire: 19.634954**
- Source de vérification : 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**
- **Rayon: 1**
- **Standard form: x² + y² = 1**
- **General form: x² + y² − 1 = 0**
- Source de vérification : 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**
- **Rayon: 1.66666667**
- **Standard form: (x − 1)² + (y − 4/3)² = 25/9**
- Source de vérification : 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.

### Quelle est la précision de « Cercle par trois points : centre, rayon et équation » ?

La précision dépend de vos données et des hypothèses de la méthode. Le calcul décimal utilise 50 chiffres significatifs, mais les estimations, méthodes numériques et données sources peuvent être moins précises ; l’arrondi affiché ne supprime pas ces limites. Exemples résolus vérifiés à partir de sources indépendantes : 4. Par exemple, « (−3, 4), (4, 5), (1, −4) » est vérifié à l’aide de Python 3.8 fractions with the determinant formula; each point is 5 from (1, 1): 4² + 3², 3² + 4², 0² + 5².

### D’où vient cette méthode ?

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

## Sources

- [Weisstein, E. W. “Circumcircle” — MathWorld (circle through three points, determinant form)](https://mathworld.wolfram.com/Circumcircle.html)
- [Weisstein, E. W. “Circle” — MathWorld (standard and general equations)](https://mathworld.wolfram.com/Circle.html)