Given two circles, of given radii, have their centres a given distance apart, such that the circles intersect each other at two points. The task is to find the length of the direct common tangent between the circles.
Examples:
Input: r1 = 4, r2 = 6, d = 3
Output: 2.23607
Input: r1 = 14, r2 = 43, d = 35
Output: 19.5959

Approach:
- Let the radii of the circles be r1 & r2 respectively.
- Let the distance between the centers be d units.
- Draw a line OR parallel to PQ
- angle OPQ = 90 deg
angle O’QP = 90 deg
{ line joining the centre of the circle to the point of contact makes an angle of 90 degrees with the tangent }
- angle OPQ + angle O’QP = 180 deg
OP || QR
- Since opposite sides are parallel and interior angles are 90, therefore OPQR is a rectangle.
- So OP = QR = r1 and PQ = OR = d
- In triangle OO’R
angle ORO’ = 90
By Pythagoras theorem
OR^2 + O’R^2 = (OO’^2)
OR^2 + (r1-r2)^2 = d^2
- so, OR^2= d^2-(r1-r2)^2
OR = √{d^2-(r1-r2)^2}

Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void lengtang( double r1, double r2, double d)
{
cout << "The length of the direct"
<< " common tangent is "
<< sqrt ( pow (d, 2) - pow ((r1 - r2), 2))
<< endl;
}
int main()
{
double r1 = 4, r2 = 6, d = 3;
lengtang(r1, r2, d);
return 0;
}
|
Java
class GFG
{
static void lengtang( double r1, double r2, double d)
{
System.out.println( "The length of the direct"
+ " common tangent is "
+ (Math.sqrt(Math.pow(d, 2 ) -
Math.pow((r1 - r2), 2 ))));
}
public static void main(String[] args)
{
double r1 = 4 , r2 = 6 , d = 3 ;
lengtang(r1, r2, d);
}
}
|
Python3
def lengtang(r1, r2, d):
print ( "The length of the direct common tangent is "
,((d * * 2 ) - ((r1 - r2) * * 2 )) * * ( 1 / 2 ));
r1 = 4 ; r2 = 6 ; d = 3 ;
lengtang(r1, r2, d);
|
C#
using System;
class GFG
{
static void lengtang( double r1, double r2, double d)
{
Console.WriteLine( "The length of the direct"
+ " common tangent is "
+ (Math.Sqrt(Math.Pow(d, 2) -
Math.Pow((r1 - r2), 2))));
}
public static void Main(String[] args)
{
double r1 = 4, r2 = 6, d = 3;
lengtang(r1, r2, d);
}
}
|
PHP
<?php
function lengtang( $r1 , $r2 , $d )
{
echo "The length of the direct common tangent is "
,sqrt(pow( $d , 2) - pow(( $r1 - $r2 ), 2)) ;
}
$r1 = 4; $r2 = 6; $d = 3;
lengtang( $r1 , $r2 , $d );
?>
|
Javascript
<script>
function lengtang(r1 , r2 , d)
{
document.write( "The length of the direct"
+ " common tangent is "
+ (Math.sqrt(Math.pow(d, 2) -
Math.pow((r1 - r2), 2))).toFixed(5));
}
var r1 = 4, r2 = 6, d = 3;
lengtang(r1, r2, d);
</script>
|
Output: The length of the direct common tangent is 2.23607
Time Complexity: O(logn) because using inbuilt sqrt and pow function
Auxiliary Space: O(1)