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Maths › Geometry › Area ›

circle

Computes the area of the circular segment within a circle tangent to a reference line.

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Interface
Circle
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Interface

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circle Computes the area of the circular segment within a circle tangent to a reference line.	<script type="text/javascript" src="http://www.codecogs.com/ic_json-393&u=3"></script>

Circle

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doublecircle(	double	`r`
	double	`h`	)[inline]

This module computes the area of the circular segment formed between a circle tangent to a reference line and a line found at a given distance from this reference line.

The various cases are described by the images given in the following documentation, where the area which we want to compute is that of the filled shape.

Solution

Consider an orthogonal coordinate system and $\mathcal{C}(O, r)$ a circle so that there exists $P \in \mathcal{C}(O, r)$ with

on the x-axis. Also let

be parallel to the x-axis so that the distance from it to the x-axis is $h \in \mathbb{R}_+$ and $d \cap \mathcal{C} = \{A, B\}$ .

Let

be the area which we must determine. Based on the relation between

and

we get the following cases:

1) $h<r \Rightarrow S = \mathcal{A}_{APB}$

We notice that

2) $h = r \Rightarrow S = \mathcal{A}_{\mathcal C} /2$

Because $\mathcal{A}_{\mathcal C} = \pi r^2$ , we find the solution:

$S = \frac{\pi r^2}{2}$

(7)

3) $r<h<2r \Rightarrow S = \mathcal{A}_{APB}$

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4) $h \geq 2r \Rightarrow S = \mathcal{A}_{\mathcal{C}}$

Because $\mathcal{A}_{\mathcal{C}} = \pi r^2$ , the solution is:

$S = \pi r^2$

(14)

Output:

r = 2.5
 
h = 0.0   Area = 0.000
h = 0.5   Area = 1.022
h = 1.0   Area = 2.796
h = 1.5   Area = 4.954
h = 2.0   Area = 7.334
h = 2.5   Area = 9.817
h = 3.0   Area = 12.301
h = 3.5   Area = 14.681
h = 4.0   Area = 16.839
h = 4.5   Area = 18.613
h = 5.0   Area = 19.635
h = 5.5   Area = 19.635

Example 1

#include <codecogs/maths/geometry/area/circle.h>
#include <stdio.h>
 
int main()
{
  // the length of the radius
  double r = 2.5;
 
  // display the lenghts of the radius
  printf("r = %.1lf\n\n", r);
 
  // display the area for different values of h
  for (double h = 0; h < 5.6; h += 0.5)
  printf("h = %.1lf   Area = %.3lf\n", h, 
  Geometry::Area::circle(r, h));
 
  return 0;
}

Parameters

r	the radius of the circle
h	the distance between line and the reference line

Returns

The value of the desired area.

Source Code

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