Given are two circles whose radii are given, such that the smaller lies completely within the bigger circle, and they touch each other at one point. We have to find the length of the perpendicular bisector of the line joining the centres of the circles.
Examples:
Input: r1 = 5, r2 = 3 Output: 9.79796 Input: r1 = 8, r2 = 4 Output: 15.4919
Approach:
- Let the two circles have center at A and B.The perpendicular bisector PQ, bisects the line at C.
- Let radius of bigger circle = r1
radius of smaller circle = r2
- so, AB = r1-r2,
- therefore, AC = (r1-r2)/2
- In, the figure, we see
PA = r1
- in triangle ACP,
PC^2 + AC^2 = PA^2
PC^2 = PA^2 – AC^2
PC^2 = r1^2 – (r1-r2)^2/4
- so, PQ = 2*√(r1^2 – (r1-r2)^2/4)
Length of the perpendicular bisector = 2 * sqrt(r1^2 – (r1-r2)*(r1-r2)/4)
Below is the implementation of the above approach:
C++
// C++ program to find the Length// of the perpendicular bisector// of the line joining the centers// of two circles in which one lies// completely inside touching the// bigger circle at one point#include <bits/stdc++.h>using namespace std;void lengperpbisect(double r1, double r2){ double z = 2 * sqrt((r1 * r1) - ((r1 - r2) * (r1 - r2) / 4)); cout << "The length of the " << "perpendicular bisector is " << z << endl;}// Driver codeint main(){ double r1 = 5, r2 = 3; lengperpbisect(r1, r2); return 0;} |
Java
// Java program to find the Length// of the perpendicular bisector// of the line joining the centers// of two circles in which one lies// completely inside touching the// bigger circle at one pointclass GFG { static void lengperpbisect(double r1, double r2){ double z = 2 * Math.sqrt((r1 * r1) - ((r1 - r2) * (r1 - r2) / 4)); System.out.println("The length of the " + "perpendicular bisector is " + z );}// Driver codepublic static void main(String[] args) { double r1 = 5, r2 = 3; lengperpbisect(r1, r2);}}// This code has been contributed by 29AjayKumar |
Python3
# Python program to find the Length# of the perpendicular bisector# of the line joining the centers# of two circles in which one lies# completely inside touching the# bigger circle at one pointdef lengperpbisect(r1, r2): z = 2 * (((r1 * r1) - ((r1 - r2) * (r1 - r2) / 4))**(1/2)); print("The length of the perpendicular bisector is ", z);# Driver coder1 = 5; r2 = 3;lengperpbisect(r1, r2);# This code contributed by PrinciRaj1992 |
C#
// C# program to find the Length// of the perpendicular bisector// of the line joining the centers// of two circles in which one lies// completely inside touching the// bigger circle at one pointusing System;class GFG { static void lengperpbisect(double r1, double r2){ double z = 2 * Math.Sqrt((r1 * r1) - ((r1 - r2) * (r1 - r2) / 4)); Console.WriteLine("The length of the " + "perpendicular bisector is " + z );}// Driver codepublic static void Main() { double r1 = 5, r2 = 3; lengperpbisect(r1, r2);}}// This code has been contributed by anuj_67.. |
Javascript
<script>// javascript program to find the Length// of the perpendicular bisector// of the line joining the centers// of two circles in which one lies// completely inside touching the// bigger circle at one pointfunction lengperpbisect(r1 , r2){ var z = 2 * Math.sqrt((r1 * r1) - ((r1 - r2) * (r1 - r2) / 4)); document.write("The length of the " + "perpendicular bisector is " + z.toFixed(5) );}// Driver codevar r1 = 5, r2 = 3;lengperpbisect(r1, r2);// This code is contributed by 29AjayKumar</script> |
Output:
The length of the perpendicular bisector is 9.79796
Time Complexity: O(log(n)), since using inbuilt sqrt function
Auxiliary Space: O(1)
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