Shortest distance from the centre of a circle to a chord

Last Updated : 22 May, 2019

Given a circle which has a chord inside it. The length of the chord and the radius of the circle are given. The task is to find the shortest distance from the chord to the centre.

Examples:

Input: r = 4, d = 3 
Output: 3.7081

Input: r = 9.8, d = 7.3
Output: 9.09492

Approach:

We know that the line segment that is dropped from the centre on the chord bisects the chord. The line is the perpendicular bisector of the chord, and we know the perpendicular distance is the shortest distance, so our task is to find the length of this perpendicular bisector.

let radius of the circle = r

length of the chord = d

so, in triangle OBC,

from pythagorus theorem,
OB^2 + (d/2)^2 = r^2
so, OB = √(r^2 – d^2/4)

So, $The shortest distance from the chord to centre = sqrt(r^2 - d^2/4)$

Below is the implementation of the above approach:

C++

// C++ program to find 
// the shortest distance from 
// chord to the centre of circle 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to find the shortest distance 
void shortdis(double r, double d) 
{ 
    cout << "The shortest distance "
         << "from the chord to centre "
         << sqrt((r * r) - ((d * d) / 4)) 
         << endl; 
} 
  
// Driver code 
int main() 
{ 
    double r = 4, d = 3; 
    shortdis(r, d); 
    return 0; 
} 

Java

// Java program to find  
// the shortest distance from  
// chord to the centre of circle  
class GFG 
{ 
      
// Function to find the shortest distance  
static void shortdis(double r, double d)  
{  
    System.out.println("The shortest distance "
        + "from the chord to centre "
        + (Math.sqrt((r * r) - ((d * d) / 4))));  
}  
  
// Driver code  
public static void main(String[] args) 
{ 
    double r = 4, d = 3;  
    shortdis(r, d);  
} 
} 
  
/* This code contributed by PrinciRaj1992 */

Python3

# Python program to find  
# the shortest distance from  
# chord to the centre of circle  
  
# Function to find the shortest distance  
def shortdis(r, d):  
    print("The shortest distance ",end=""); 
    print("from the chord to centre ",end=""); 
    print(((r * r) - ((d * d) / 4))**(1/2)); 
  
# Driver code  
r = 4; 
d = 3;  
shortdis(r, d);  
  
# This code has been contributed by 29AjayKumar 

C#

// C# program to find  
// the shortest distance from  
// chord to the centre of circle  
using System; 
  
class GFG  
{  
      
    // Function to find the shortest distance  
    static void shortdis(double r, double d)  
    {  
        Console.WriteLine("The shortest distance "
            + "from the chord to centre "
            + (Math.Sqrt((r * r) - ((d * d) / 4))));  
    }  
      
    // Driver code  
    public static void Main()  
    {  
        double r = 4, d = 3;  
        shortdis(r, d);  
    }  
}  
  
// This code is contributed by AnkitRai01 

PHP

<?php 
  
// PHP program to find  
// the shortest distance from  
// chord to the centre of circle  
  
  
// Function to find the shortest distance  
function shortdis($r,  $d)  
{  
    echo "The shortest distance "; 
    echo "from the chord to centre "; 
    echo sqrt(($r * $r) - (($d * $d) / 4)); 
}  
  
// Driver code  
    $r = 4; 
    $d = 3;  
    shortdis($r, $d);  
  
// This code is contributed by Naman_Garg. 
  
?> 

Output:

The leshortest distance from the chord to centre 3.7081

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