We wish to estimate the area of the intersection of two overlapping circles with

Question

We wish to estimate the area of the intersection of two overlapping circles with diameters dl and d2, whose centres are separated by z. We can do this by a simple sampling procedure. Imagine the two circles arc enclosed in a rectangle with width d1+d2 and height dk, where dk is the greater of dl and d2: [Dl Generate n points randomly in this rectangle, and count how many points sit in the intersection (i.e. inside both circles). The proportion of points that sit in the intersection approximates A/B, where A is the area of the intersection, and B is the area of the rectangle. From this we can estimate A. Clearly the precision of the estimate increases with the number of samples. Write a method public double area(double dl, double d2, double x, int n) that implements the above procedure. You should use an object from the library class java .until .Random to generate a sequence of random points inside the rectangle, and you will need the following method from that class. public double nextDouble() Returns the next pseudorandom, uniformly-distributed double between 0.0 and 1.0 from this random number generator?s sequence.

Explanation / Answer

import java.util.Random;

import java.util.Scanner;

public class AreaOfIntersection {

public double area(double d1, double d2, double x, int n)
{
  
double width=d1+d2;
double hight=0;
double []xc=new double[n]; // to store x-coordinate of random point
double []yc=new double[n]; // to store y-coordinate of random point
Random r=new Random(); // object of random class
  
  
double A=0;// area of intersection
  
if(d1>d2)
{
hight=d1;
}
else
hight=d2;
// Area of rectangle is
double B=width*hight;
// generating points
for(int i=0;i<n;i++)
{
xc[i]=r.nextDouble()*10;
yc[i]=r.nextDouble()*10;
}
  
// let center of the circle (0,0) and (x,0)
int count=0;

for (int i=0;i<n;i++)
{
boolean temp1=false,temp2=false;
// whether point lies inside or out side the circle
// using x^2+y^2<r^2
if(((yc[i]-d1/2)*(yc[i]-d1/2)+(xc[i]-d1/2)*(xc[i]-d1/2))<((d1/2)*(d1/2)))
temp1=true;
if(((yc[i]-d1/2)*(yc[i]-d1/2)+(xc[i]-(x+d1/2))*(xc[i]-(x+d1/2)))<((d2/2)*(d2/2)))
temp2=true;
if(temp1&&temp2)
{
count++; // Cunting the points which lies inside the intersection
}
}
  
// as count is proportional to A/B
// so A=count*B;
A=count*B;
return A;
  
}
public static void main(String[] args) {
  
Scanner sc=new Scanner(System.in);
// reading input
double d1=sc.nextDouble();
double d2=sc.nextDouble();
double x=sc.nextDouble();
int n=sc.nextInt();
double A=0;
// creating instance of method
AreaOfIntersection ob=new AreaOfIntersection();
A=ob.area(d1, d2, x, n);
System.out.println(A);
  
}
}

output: (According to given in question Count is Approximates to (A/B) .....so A=count*B.......where A is the intersection area.....if I interpreted wrong then answer may vary .....please make change only in formula accordingly )

d1=5
d2=4
x=2
n=100
A=360.0

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