# 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 = 4
Output: 50
Explanation:
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 = 4
Output: 36

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

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

1. The number of ways to choose a pair of points from X circles is . Each such pair intersect at most two points.
2. The number of ways to choose a pair of points from Y lines is . Each such pair intersect in at most one point.
3. 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 of the above approach  #include  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

Output:

36


Time Complexity: O(1)
Auxiliary Space: O(1) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : princi singh, btc_148