Find N random points within a Circle

Given four integers N, R, X, and Y such that it represents a circle of radius R with [X, Y] as coordinates of the center. The task is to find N random points inside or on the circle.
Examples:

Input: R = 12, X = 3, Y = 3, N = 5
Output: (7.05, -3.36) (5.21, -7.49) (7.53, 0.19) (-2.37, 12.05) (1.45, 11.80)

Input: R = 5, X = 1, Y = 1, N = 3
Output: (4.75, 1.03) (2.57, 5.21) (-1.98, -0.76)

Approach: To find a random point in or on a circle we need two components, an angle(theta) and distance(D) from the center. After that Now, the point (x_i, y_i) can be expressed as:

x_i = X + D * cos(theta)
y_i = Y + D * sin(theta)

Below is the implementation of the above approach:

C++

// C++ program for the above approach 
#include <bits/stdc++.h> 
using namespace std; 
#define PI 3.141592653589 
  
// Return a random double between 0 & 1 
double uniform() 
{ 
    return (double)rand() / RAND_MAX; 
} 
  
// Function to find the N random points on 
// the given circle 
vector<pair<double, double> > randPoint( 
    int r, int x, int y, int n) 
{ 
    // Result vector 
    vector<pair<double, double> > res; 
  
    for (int i = 0; i < n; i++) { 
  
        // Get Angle in radians 
        double theta = 2 * PI * uniform(); 
  
        // Get length from center 
        double len = sqrt(uniform()) * r; 
  
        // Add point to results. 
        res.push_back({ x + len * cos(theta), 
                        y + len * sin(theta) }); 
    } 
  
    // Return the N points 
    return res; 
} 
  
// Function to display the content of 
// the vector A 
void printVector( 
    vector<pair<double, double> > A) 
{ 
  
    // Iterate over A 
    for (pair<double, double> P : A) { 
  
        // Print the N random points stored 
        printf("(%.2lf, %.2lf)\n", 
               P.first, P.second); 
    } 
} 
  
// Driver Code 
int main() 
{ 
    // Given dimensions 
    int R = 12; 
    int X = 3; 
    int Y = 3; 
    int N = 5; 
  
    // Function Call 
    printVector(randPoint(R, X, Y, N)); 
    return 0; 
} 

Java

// Java program for the above approach 
import java.util.*; 
  
class GFG{ 
      
static final double PI = 3.141592653589; 
static class pair  
{ 
    double first, second; 
  
    public pair(double first, 
                double second) 
    { 
        super(); 
        this.first = first; 
        this.second = second; 
    } 
} 
  
// Return a random double between 0 & 1 
static double uniform(){return Math.random();} 
  
// Function to find the N random points on 
// the given circle 
static Vector<pair> randPoint(int r, int x,  
                              int y, int n) 
{ 
      
    // Result vector 
    Vector<pair> res = new Vector<pair>(); 
  
    for(int i = 0; i < n; i++) 
    { 
          
        // Get Angle in radians 
        double theta = 2 * PI * uniform(); 
  
        // Get length from center 
        double len = Math.sqrt(uniform()) * r; 
  
        // Add point to results. 
        res.add(new pair(x + len * Math.cos(theta), 
                         y + len * Math.sin(theta))); 
    } 
      
    // Return the N points 
    return res; 
} 
  
// Function to display the content of 
// the vector A 
static void printVector(Vector<pair> A) 
{ 
  
    // Iterate over A 
    for(pair P : A) 
    { 
          
        // Print the N random points stored 
        System.out.printf("(%.2f, %.2f)\n",  
                          P.first, P.second); 
    } 
} 
  
// Driver Code 
public static void main(String[] args) 
{ 
      
    // Given dimensions 
    int R = 12; 
    int X = 3; 
    int Y = 3; 
    int N = 5; 
  
    // Function call 
    printVector(randPoint(R, X, Y, N)); 
} 
} 
  
// This code is contributed by Rajput-Ji 

Output:

(7.05, -3.36)
(5.21, -7.49)
(7.53, 0.19)
(-2.37, 12.05)
(1.45, 11.80)

Time Complexity: O(N)
Space Complexity: O(N)

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My Personal Notes

C++

Java

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