圆心到弦的最短距离

给定一个圆圈，里面有一个和弦。给出了弦的长度和圆的半径。任务是找到和弦到中心的最短距离。
例子：

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

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

方法：

• 我们知道从和弦的中心放下的线段将和弦二等分。该线是弦的垂直平分线，我们知道垂直距离是最短的距离，所以我们的任务是找到此垂直平分线的长度。

• 令圆的半径= r

• 弦长= d

• 因此，在OBC三角形中，
  根据毕达哥拉斯定理，
  OB ^ 2 +(d / 2)^ 2 = r ^ 2
  因此， OB =√(r ^ 2 – d ^ 2/4)

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

下面是上述方法的实现：

C++

// C++ program to find
// the shortest distance from
// chord to the centre of circle
#include 
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

Javascript

 
// JavaScript program to find
// the shortest distance from
// chord to the centre of circle
 
// Function to find the shortest distance
function shortdis(r, d)
{
    document.write("The shortest distance "
        + "from the chord to centre "
        + Math.sqrt((r * r) - ((d * d) / 4))
        + "<br>");
}
 
// Driver code
 
    let r = 4, d = 3;
    shortdis(r, d);
 
// This code is contributed by Surbhi Tyagi.
 

输出：

The leshortest distance from the chord to centre 3.7081