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Lingkaran melalui tiga titik: pusat, jari-jari, persamaan

Tentukan pusat, jari-jari, luas dan persamaan lingkaran melalui tiga titik, dalam bentuk (x − h)² + (y − k)² = r² dan bentuk umum.

Diperbarui Contoh terverifikasi: 4

Coba
Jari-jari
Jari-jari: 5
Tempat desimal maksimum: 8; Terdekat, jika berjarak sama menjauhi nol
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
Keliling lingkaran
31.4159
Luas
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)
Cara menghitung 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. Jari-jari

    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

Tentang Lingkaran melalui tiga titik: pusat, jari-jari, persamaan

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.

Contoh penyelesaian

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

Sumber pemeriksaan: 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
Jari-jari
2.5
Luas
19.634954

Sumber pemeriksaan: 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
Jari-jari
1
Standard form
x² + y² = 1
General form
x² + y² − 1 = 0

Sumber pemeriksaan: 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
Jari-jari
1.66666667
Standard form
(x − 1)² + (y − 4/3)² = 25/9

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

Pertanyaan

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.

Seberapa akurat “Lingkaran melalui tiga titik: pusat, jari-jari, persamaan”?

Akurasi bergantung pada masukan dan asumsi metode. Aritmetika desimal memakai 50 digit signifikan, tetapi perkiraan, metode numerik dan data sumber bisa kurang presisi; pembulatan yang ditampilkan tidak menghilangkan batasan itu. Contoh penyelesaian yang diperiksa dengan sumber independen: 4. Misalnya, “(−3, 4), (4, 5), (1, −4)” diperiksa dengan Python 3.8 fractions with the determinant formula; each point is 5 from (1, 1): 4² + 3², 3² + 4², 0² + 5².

Dari mana metode ini berasal?

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

Tentang kalkulator ini

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}

Sumber

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

Diperiksa dengan referensi

Kalkulator ini mencakup 4 contoh perhitungan dengan jawaban dari sumber independen. Contoh tersebut dijalankan dalam rangkaian pengujian dan dapat Anda jalankan di sini juga.

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