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# Circle and Lattice Points

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 x2 + y2 = r2
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 the implementation of above idea.

## CPP

 `// C++ program to find countLattice points on a circle``#include``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

## 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

## Python3

 `# Python3 program to find``# countLattice points on a circle` `import` `math` `# Function to count Lattice``# points 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 integer``        ``# 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`

## C#

 `// C# program to find countLattice``// points on a circle``using` `System;` `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()``    ``{``        ``int` `r = 5;``        ` `        ``Console.Write(countLattice(r));``    ``}``}` `// This code is contributed by nitin mittal.`

## PHP

 ``

## Javascript

 ``

Output:

`12`

Time Complexity: O(1)
Auxiliary Space: O(1)

Reference:
http://mathworld.wolfram.com/CircleLatticePoints.html
This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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