# 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

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 . 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  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 import java.io.*; public 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

 

Output:

36

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

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