Given a circle of radius **r** in 2-D with origin or (0, 0) as center. The task is to find the total lattice points on circumference. Lattice Points are points with coordinates as integers in 2-D space.

Example:

Input : r = 5. Output : 12 Below are lattice points on a circle with radius 5 and origin as (0, 0). (0,5), (0,-5), (5,0), (-5,0), (3,4), (-3,4), (-3,-4), (3,-4), (4,3), (-4,3), (-4,-3), (4,-3). are 12 lattice point.

To find lattice points, we basically need to find values of (x, y) which satisfy the equation x^{2} + y^{2} = r^{2}.

For any value of (x, y) that satisfies the above equation we actually have total 4 different combination which that satisfy the equation. For example if r = 5 and (3, 4) is a pair which satisfies the equation, there are actually 4 combinations (3, 4) , (-3,4) , (-3,-4) , (3,-4). There is an exception though, in case of (0, r) or (r, 0) there are actually 2 points as there is no negative 0.

// Initialize result as 4 for (r, 0), (-r. 0), // (0, r) and (0, -r) result = 4 Loop for x = 1 to r-1 and do following for every x. If r*r - x*x is a perfect square, then add 4 tor result.

Below is C++ implementation of above idea.

## CPP

// C++ program to find countLattice points on a circle #include<bits/stdc++.h> using namespace std; // Function to count Lattice points on a circle int countLattice(int r) { if (r <= 0) return 0; // Initialize result as 4 for (r, 0), (-r. 0), // (0, r) and (0, -r) int result = 4; // Check every value that can be potential x for (int x=1; x<r; x++) { // Find a potential y int ySquare = r*r - x*x; int y = sqrt(ySquare); // checking whether square root is an integer // or not. Count increments by 4 for four // different quadrant values if (y*y == ySquare) result += 4; } return result; } // Driver program int main() { int r = 5; cout << countLattice(r); return 0; }

## Java

// Java program to find // countLattice points on a circle class GFG { // Function to count // Lattice points on a circle static int countLattice(int r) { if (r <= 0) return 0; // Initialize result as 4 for (r, 0), (-r. 0), // (0, r) and (0, -r) int result = 4; // Check every value that can be potential x for (int x=1; x<r; x++) { // Find a potential y int ySquare = r*r - x*x; int y = (int)Math.sqrt(ySquare); // checking whether square root is an integer // or not. Count increments by 4 for four // different quadrant values if (y*y == ySquare) result += 4; } return result; } // Driver code public static void main(String arg[]) { int r = 5; System.out.println(countLattice(r)); } } // This code is contributed by Anant Agarwal.

## Python3

# Python3 program to find # countLattice podefs on a circle import math # Function to count Lattice # podefs on a circle def countLattice(r): if (r <= 0): return 0 # Initialize result as 4 for (r, 0), (-r. 0), # (0, r) and (0, -r) result = 4 # Check every value that can be potential x for x in range(1, r): # Find a potential y ySquare = r*r - x*x y = int(math.sqrt(ySquare)) # checking whether square root is an defeger # or not. Count increments by 4 for four # different quadrant values if (y*y == ySquare): result += 4 return result # Driver program r = 5 print(countLattice(r)) # This code is contributed by # Smitha Dinesh Semwal

Output:

12

**Reference:**

http://mathworld.wolfram.com/CircleLatticePoints.html

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