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 <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
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

<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)