# Find minimum radius such that atleast k point lie inside the circle

Given a positive integer K, a circle center at (0, 0) and coordinates of some points. The task is to find minimum radius of the circle so that at-least k points lie inside the circle. Output the square of the minimum radius.

Examples:

```Input : (1, 1), (-1, -1), (1, -1),
k = 3
Output : 2
We need a circle of radius at least 2
to include 3 points.

Input : (1, 1), (0, 1), (1, -1),
k = 2
Output : 1
We need a circle of radius at least 1
to include 2 points. The circle around
(0, 0) of radius 1 would include (1, 1)
and (0, 1).
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to find square of Euclidean Distance of each point from origin (0, 0). Now, sort these distance in increasing order. Now the kth element of distance is the required minimum radius.

Below is the implementation of this approach:

## C++

 `// C++ program to find minimum radius  ` `// such that atleast k point lie inside ` `// the circle ` `#include ` `using` `namespace` `std; ` ` `  `// Return minumum distance required so that ` `// aleast k point lie inside the circle. ` `int` `minRadius(``int` `k, ``int` `x[], ``int` `y[], ``int` `n) ` `{ ` `   ``int` `dis[n]; ` `     `  `   ``// Finding distance between of each ` `   ``// point from origin ` `   ``for` `(``int` `i = 0; i < n; i++) ` `       ``dis[i] = x[i] * x[i] + y[i] * y[i]; ` `     `  `    ``// Sorting the distance ` `    ``sort(dis, dis + n); ` `     `  `    ``return` `dis[k - 1]; ` `} ` ` `  `// Driven Program ` `int` `main() ` `{ ` `  ``int` `k = 3; ` `  ``int` `x[] = { 1, -1, 1 }; ` `  ``int` `y[] = { 1, -1, -1 }; ` `  ``int` `n = ``sizeof``(x)/``sizeof``(x); ` `     `  `  ``cout << minRadius(k, x, y, n) << endl; ` `     `  `  ``return` `0; ` `} `

## Java

 `// Java program to find minimum radius  ` `// such that atleast k point lie inside ` `// the circle ` `import` `java.util.Arrays; ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Return minumum distance required so that  ` `    ``// aleast k point lie inside the circle. ` `    ``static` `int` `minRadius(``int` `k, ``int``[] x, ``int``[] y,  ` `                                          ``int` `n) ` `    ``{ ` `        ``int``[] dis=``new` `int``[n]; ` `     `  `        ``// Finding distance between of each ` `        ``// point from origin ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `            ``dis[i] = x[i] * x[i] + y[i] * y[i]; ` `     `  `        ``// Sorting the distance ` `        ``Arrays.sort(dis); ` `     `  `        ``return` `dis[k - ``1``]; ` `    ``} ` ` `  `    ``// Driven Program ` `    ``public` `static` `void` `main (String[] args) { ` `         `  `    ``int` `k = ``3``; ` `    ``int``[] x = { ``1``, -``1``, ``1` `}; ` `    ``int``[] y = { ``1``, -``1``, -``1` `}; ` `    ``int` `n = x.length; ` `     `  `    ``System.out.println(minRadius(k, x, y, n));  ` ` `  `    ``} ` `} ` ` `  `/* This code is contributed by Mr. Somesh Awasthi */`

## Python3

 `# Python3 program to find minimum radius  ` `# such that atleast k point lie inside ` `# the circle ` ` `  ` `  `# Return minumum distance required so  ` `# that aleast k point lie inside the  ` `# circle. ` `def` `minRadius(k, x, y, n): ` `    ``dis ``=` `[``0``] ``*` `n ` ` `  `    ``# Finding distance between of each ` `    ``# point from origin ` ` `  `    ``for` `i ``in` `range``(``0``, n): ` `        ``dis[i] ``=` `x[i] ``*` `x[i] ``+` `y[i] ``*` `y[i] ` ` `  `    ``# Sorting the distance ` `    ``dis.sort() ` ` `  `    ``return` `dis[k ``-` `1``] ` `         `  `# Driver Program ` `k ``=` `3` `x ``=` `[``1``, ``-``1``, ``1``] ` `y ``=` `[``1``, ``-``1``, ``-``1``] ` `n ``=` `len``(x) ` ` `  `print``(minRadius(k, x, y, n)) ` ` `  `# This code is contributed by ` `# Prasad Kshirsagar `

## C#

 `// C# program to find minimum radius  ` `// such that atleast k point lie inside ` `// the circle ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// Return minumum distance required ` `    ``// so that aleast k point lie inside ` `    ``// the circle. ` `    ``static` `int` `minRadius(``int` `k, ``int` `[]x, ` `                          ``int``[] y, ``int` `n) ` `    ``{ ` `        ``int``[] dis = ``new` `int``[n]; ` `     `  `        ``// Finding distance between of ` `        ``// each point from origin ` `        ``for` `(``int` `i = 0; i < n; i++) ` `            ``dis[i] = x[i] * x[i] + ` `                       ``y[i] * y[i]; ` `     `  `        ``// Sorting the distance ` `        ``Array.Sort(dis); ` `     `  `        ``return` `dis[k - 1]; ` `    ``} ` ` `  `    ``// Driven Program ` `    ``public` `static` `void` `Main () ` `    ``{ ` `        ``int` `k = 3; ` `        ``int``[] x = { 1, -1, 1 }; ` `        ``int``[] y = { 1, -1, -1 }; ` `        ``int` `n = x.Length; ` `         `  `        ``Console.WriteLine( ` `              ``minRadius(k, x, y, n));  ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output:

```2
```

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