# Üç noktadan çember: merkez, yarıçap ve denklem

> Üç noktadan geçen çemberin merkezini, yarıçapını, alanını ve denklemini (x − h)² + (y − k)² = r² biçiminde ve genel biçimde bulun.

Etkileşimli sürüm: https://www.calcopenly.com/tr/geometry/circle-through-three-points
Konu: Geometri hesaplayıcıları

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.

## Girdiler

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

## Sonuçlar

- Yarıçap — ana sonuç
- Centre x
- Centre y
- Standard form
- General form
- Çap
- Çember çevresi
- Alan

## Formül

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

## Çözümlü örnekler

### (−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**
- **Yarıçap: 5**
- **Standard form: (x − 1)² + (y − 1)² = 25**
- **General form: x² + y² − 2x − 2y − 23 = 0**
- Doğrulama kaynağı: 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**
- **Yarıçap: 2.5**
- **Alan: 19.634954**
- Doğrulama kaynağı: 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**
- **Yarıçap: 1**
- **Standard form: x² + y² = 1**
- **General form: x² + y² − 1 = 0**
- Doğrulama kaynağı: 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**
- **Yarıçap: 1.66666667**
- **Standard form: (x − 1)² + (y − 4/3)² = 25/9**
- Doğrulama kaynağı: Python 3.8 fractions: h = 1 by symmetry, 9 − 6k = 1 gives k = 4/3, r² = 1 + 16/9 = 25/9

## Sorular

### 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.

### “Üç noktadan çember: merkez, yarıçap ve denklem” ne kadar doğru sonuç verir?

Doğruluk, girdilerinize ve yöntemin varsayımlarına bağlıdır. Ondalık aritmetik 50 anlamlı basamak kullanır; ancak tahminler, sayısal yöntemler ve kaynak veriler daha az hassas olabilir. Gösterilen değerin yuvarlanması bu sınırları ortadan kaldırmaz. Bağımsız kaynaklarla doğrulanan çözümlü örnek sayısı: 4. Örneğin “(−3, 4), (4, 5), (1, −4)”, Python 3.8 fractions with the determinant formula; each point is 5 from (1, 1): 4² + 3², 3² + 4², 0² + 5² ile karşılaştırılarak doğrulanır.

### Yöntemin kaynağı nedir?

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

## Kaynaklar

- [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)