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세 점을 지나는 원: 중심, 반지름과 방정식

세 점을 지나는 원의 중심, 반지름, 넓이와 방정식을 구하세요. 표준형 (x − h)² + (y − k)² = r²과 일반형을 제공합니다.

업데이트 검증한 예제: 4

시도하기
반지름
반지름: 5
최대 소수 자릿수: 8; 가장 가까운 값, 중간값은 0에서 먼 쪽으로
Centre x
1
Centre y
1
Standard form
(x − 1)² + (y − 1)² = 25
General form
x² + y² − 2x − 2y − 23 = 0
지름
10
원둘레
31.4159
넓이
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

r = 5centre (1, 1)P1(−3, 4)P2(4, 5)P3(1, −4)
계산 방법 S

  1. Determinant (zero means the points are collinear)

    D=2[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]=−120D = 2[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] = -120

  2. Centre

    h=(x12+y12)(y2−y3)+(x22+y22)(y3−y1)+(x32+y32)(y1−y2)D=1k=(x12+y12)(x3−x2)+(x22+y22)(x1−x3)+(x32+y32)(x2−x1)D=1\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.

  3. 반지름

    r=(x1−h)2+(y1−k)2=25=5r = \sqrt{(x_1 - h)^2 + (y_1 - k)^2} = \sqrt{25} = 5

  4. Equation

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

세 점을 지나는 원: 중심, 반지름과 방정식 소개

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.

계산 예제

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

검증 출처: 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
반지름
2.5
넓이
19.634954

검증 출처: 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
반지름
1
Standard form
x² + y² = 1
General form
x² + y² − 1 = 0

검증 출처: 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
반지름
1.66666667
Standard form
(x − 1)² + (y − 4/3)² = 25/9

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

자주 묻는 질문

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.

“세 점을 지나는 원: 중심, 반지름과 방정식”의 정확도는 어느 정도인가요?

정확도는 입력값과 계산 방법의 가정에 따라 달라집니다. 십진 연산은 유효숫자 50자리를 사용하지만, 추정값·수치해석 방법·원본 데이터의 정밀도는 더 낮을 수 있습니다. 표시값을 반올림해도 이러한 한계는 사라지지 않습니다. 독립적인 출처의 풀이와 대조한 계산 예시: 4. 예를 들어 “(−3, 4), (4, 5), (1, −4)”은 Python 3.8 fractions with the determinant formula; each point is 5 from (1, 1): 4² + 3², 3² + 4², 0² + 5²와 대조해 확인합니다.

이 계산 방법의 출처는 무엇인가요?

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

이 계산기 소개

h=∑(xi2+yi2)(yj−yk)2∑xi(yj−yk),k=∑(xi2+yi2)(xk−xj)2∑xi(yj−yk)(x−h)2+(y−k)2=r2\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}

출처

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

출처와 대조하여 검증

이 계산기에는 독립적인 출처에서 답을 얻은 계산 예제가 4개 있습니다. 테스트 모음에서 실행되며 여기에서도 실행할 수 있습니다.

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