Окружность по трём точкам: центр, радиус и уравнение

Найдите центр, радиус, площадь и уравнение окружности через три точки, в форме (x − h)² + (y − k)² = r² и в общем виде.

Обновлено 28 сентября 2026 г.Проверенные примеры: 4

Радиус

Максимум знаков после запятой: 8; До ближайшего, при равенстве — от нуля

Centre x: 1
Centre y: 1
Standard form: (x − 1)² + (y − 1)² = 25
General form: x² + y² − 2x − 2y − 23 = 0
Диаметр: 10
Длина окружности: 31.4159
Площадь: 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

Как выполняется расчёт 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.
Радиус
$r = \sqrt{(x_1 - h)^2 + (y_1 - k)^2} = \sqrt{25} = 5$
Equation
Standard: (x − 1)² + (y − 1)² = 25 · General: x² + y² − 2x − 2y − 23 = 0

О калькуляторе: Окружность по трём точкам: центр, радиус и уравнение

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.

Примеры с решением

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

Источник проверки: 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
Радиус: 2.5
Площадь: 19.634954

Источник проверки: 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
Радиус: 1
Standard form: x² + y² = 1
General form: x² + y² − 1 = 0

Источник проверки: 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
Радиус: 1.66666667
Standard form: (x − 1)² + (y − 4/3)² = 25/9

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

Вопросы

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.

Насколько точен «Окружность по трём точкам: центр, радиус и уравнение»?

Точность зависит от введённых данных и допущений метода. Десятичная арифметика использует 50 значащих цифр, но оценки, численные методы и исходные данные могут быть менее точными; округление на экране не устраняет эти ограничения. Решённые примеры, проверенные по независимым источникам: 4. Например, «(−3, 4), (4, 5), (1, −4)» проверяется по источнику Python 3.8 fractions with the determinant formula; each point is 5 from (1, 1): 4² + 3², 3² + 4², 0² + 5².

Откуда взята методика?

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

Об этом калькуляторе

\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}

Источники

Проверено по источникам

В калькулятор включены решённые примеры с ответами из независимых источников. Их количество: 4. Они входят в набор тестов, и вы также можете запустить их здесь.