# Maximum points of intersections possible among X circles and Y straight lines

Given two integers **X** and **Y**, the task is to find the maximum number of points of intersection possible among **X** circles and **Y** straight lines.

**Example:**

Input:X = 4, Y = 4Output:50Explanation:

4 lines intersect each other at 6 points and 4 circles intersect each other at maximum of 12 points.

Each line intersects 4 circles at 8 points.

Hence, 4 lines intersect four circles at a maximum of 32 points.

Thus, required number of intersections = 6 + 12 + 32 = 50.

Input:X = 3, Y = 4Output:36

**Approach:**

It can be observed that there are three types of intersections:

- The number of ways to choose a pair of points from X circles is . Each such pair intersect at most two points.
- The number of ways to choose a pair of points from Y lines is . Each such pair intersect in at most one point.
- The number of ways to choose one circle and one line from X circles and Y lines is is . Each such pair intersect in at most two points.

So, the maximum number of point of intersection can be calculated as:

=>

=>

Thus, formula to find maximum number of point of intersection of **X** circles and **Y** straight lines is:

Below is the implementation of the above approach:

## C++

`// C++ program to implement` `// the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `int` `maxPointOfIntersection(` `int` `x, ` `int` `y)` `{` ` ` `int` `k = y * (y - 1) / 2;` ` ` `k = k + x * (2 * y + x - 1);` ` ` `return` `k;` `}` `// Driver code` `int` `main()` `{` ` ` ` ` `// Number of circles` ` ` `int` `x = 3;` ` ` ` ` `// Number of straight lines` ` ` `int` `y = 4;` ` ` `// Function Call` ` ` `cout << (maxPointOfIntersection(x, y));` `}` `// This code is contributed by Ritik Bansal` |

## Java

`// Java program to implement` `// the above approach` `class` `GFG{` ` ` `static` `int` `maxPointOfIntersection(` `int` `x, ` `int` `y)` `{` ` ` `int` `k = y * (y - ` `1` `) / ` `2` `;` ` ` `k = k + x * (` `2` `* y + x - ` `1` `);` ` ` `return` `k;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` ` ` `// Number of circles` ` ` `int` `x = ` `3` `;` ` ` ` ` `// Number of straight lines` ` ` `int` `y = ` `4` `;` ` ` `// Function Call` ` ` `System.out.print(maxPointOfIntersection(x, y));` `}` `}` `// This code is contributed by Princi Singh` |

## Python3

`# Python3 program to implement` `# the above approach` `def` `maxPointOfIntersection(x, y):` ` ` `k ` `=` `y ` `*` `( y ` `-` `1` `) ` `/` `/` `2` ` ` `k ` `=` `k ` `+` `x ` `*` `( ` `2` `*` `y ` `+` `x ` `-` `1` `)` ` ` `return` `k` `# Number of circles` `x ` `=` `3` `# Number of straight lines` `y ` `=` `4` `# Function Call` `print` `(maxPointOfIntersection(x, y))` |

## C#

`// C# program to implement` `// the above approach` `using` `System;` `class` `GFG{` ` ` `static` `int` `maxPointOfIntersection(` `int` `x, ` `int` `y)` `{` ` ` `int` `k = y * (y - 1) / 2;` ` ` `k = k + x * (2 * y + x - 1);` ` ` `return` `k;` `}` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` ` ` `// Number of circles` ` ` `int` `x = 3;` ` ` ` ` `// Number of straight lines` ` ` `int` `y = 4;` ` ` `// Function Call` ` ` `Console.Write(maxPointOfIntersection(x, y));` `}` `}` `// This code is contributed by Princi Singh` |

## Javascript

`<script>` `// Javascript program to implement` `// the above approach` `function` `maxPointOfIntersection(x, y)` `{` ` ` `let k = y * (y - 1) / 2;` ` ` `k = k + x * (2 * y + x - 1);` ` ` `return` `k;` `}` `// Driver code` `// Number of circles` `let x = 3;` ` ` `// Number of straight lines` `let y = 4;` `// Function Call` `document.write(maxPointOfIntersection(x, y));` `// This code is contributed by rameshtravel07` `</script>` |

**Output:**

36

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