Check whether a point exists in circle sector or not.

We have a circle centered at origin (0, 0). As input we are given with starting angle of the circle sector and the size of the circle sector in percentage.
Examples:

Input :  Radius = 8 
         StartAngle = 0 
         Percentage = 12 
         x = 3 y = 4 
Output : Point (3, 4) exists in the circle 
         sector

Input : Radius = 12 
        Startangle = 45
        Percentage = 25  
        x = 3 y = 4 
Output : Point (3, 4) does not exist in 
         the circle sector
Source:wikibooks.org

Source:wikibooks.org

In this image starting angle is 0 degree, radius r and suppose that percentage of colored area is 12% then we calculate Ending Angle as 360/percentage + starting angle.

To find whether a point (x, y) exists in a circle sector (centered at origin) or not we find polar coordinates of that point and then go through the following steps:

  1. Convert x, y to polar coordinates using this
    Angle = atan(y/x); Radius = sqrt(x * x + y * y);
  2. Then Angle must be between StartingAngle and EndingAngle, and Radius between 0 and your Radius.
// C++ program to check if a point lies inside a circle
// sector.
#include<bits/stdc++.h>
using namespace std;

void checkPoint(int radius, int x, int y, float percent,
                                         float startAngle)
{
    // calculate endAngle
    float endAngle = 360/percent + startAngle;

    // Calculate polar co-ordinates
    float polarradius = sqrt(x*x+y*y);
    float Angle = atan(y/x);

    // Check whether polarradius is less then radius of circle
    // or not and Angle is between startAngle and endAngle
    // or not
    if (Angle>=startAngle && Angle<=endAngle && polarradius<radius)
        printf("Point (%d, %d) exist in the circle sector\n", x, y);
    else
        printf("Point (%d, %d) does not exist in the circle sector\n",
                                                              x, y);
}

// Driver code
int main()
{
    int radius = 8, x = 3, y = 4;
    float percent  = 12, startAngle = 0;
    checkPoint(radius, x, y, percent, startAngle);
    return 0;
}

Output: Point(3, 4) exists in the circle sector

Time complexity = O(1)

This article is contributed by Niteesh kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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