Check if a given circle lies completely inside the ring formed by two concentric circles

Given two circles of radius r and R, both have their centre at the origin. Now, given another circle of radius r1 and centre at (x1, y1). Check, if the third circle(circle of radius r1) lies completely inside the ring formed by two circles of radius r and R.
Examples : 

Input : r = 8 R = 4
        r1 = 2 x1 = 6 y1 = 0
Output : yes

Input : r = 8 R = 4 
        r1 = 2 x1 = 5 y1 = 0
Output : no

Important : Concentric circles are those circles which have same centre. The region lying between two concentric circles is called annulus or the circular ring. 

Example :
There are two concentric circles with their centre at origin(0, 0) and radius as r = 8 and R = 4. 
1.) Circle 1 and 2 lies inside the ring. 
2.) Circle 3 and 4 are outside the ring.
The complete figure can be visualised as given below : 

Approach :
This problem can be solved using Pythagoras Theorem . Compute the distance between the centre of the circle and origin using Pythagoras theorem, suppose it is denoted by ‘dis’. 
After computing the distance just check that the value of (dis – r1)> = r and (dis + r1)< = R. If both these conditions hold then the circle lies completely inside the ring.

C++

// CPP code to check if a circle
// lies in the ring
#include <bits/stdc++.h>
using namespace std;

// Function to check if circle
// lies in the ring
bool checkcircle(int r, int R, int r1,
                        int x1, int y1)
{
    // distance between center of circle
    // center of concentric circles(origin)
    // using Pythagoras theorem
    int dis = sqrt(x1*x1+y1*y1);

    // Condition to check if circle is
    // strictly inside the ring
    return (dis-r1 >= R && dis+r1 <= r);
}

// Driver Code
int main()
{
    // Both circle with radius 'r'
    // and 'R' have center (0,0)
    int r = 8, R = 4, r1 = 2, x1 = 6, y1 = 0;
    if (checkcircle(r, R, r1, x1, y1))
       cout << "yes" << endl;
    else
       cout << "no" << endl;

    return 0;
}

Java

// Java code to check if a
// circle lies in the ring
import java.io.*;

class ring
{
    // Function to check if circle
    // lies in the ring
    public static boolean checkcircle(int r, int R,
                            int r1, int x1, int y1)
    {
        // distance between center of circle
        // center of concentric circles(origin)
        // using Pythagoras theorem
        int dis = (int)Math.sqrt(x1 * x1 +
                                 y1 * y1);

         // Condition to check if circle
         // is strictly inside the ring
        return (dis - r1 >= R && dis + r1 <= r);
    }

    // Driver Code
    public static void main(String args[])
    {
        // Both circle with radius 'r'
        // and 'R' have center (0,0)
        int r = 8, R = 4, r1 = 2, x1 = 6, y1 = 0;

        if (checkcircle(r, R, r1, x1, y1))
            System.out.println("yes");
        else
            System.out.println("no");
    }
}

Python3

# Python3 code to check if
# a circle  lies in the ring
import math

# Function to check if circle
# lies in the ring
def checkcircle(r, R, r1, x1, y1):

    # distance between center of circle
    # center of concentric circles(origin)
    # using Pythagoras theorem
    dis = int(math.sqrt(x1 * x1 + y1 * y1))

    # Condition to check if circle is
    # strictly inside the ring
    return (dis-r1 >= R and dis+r1 <= r)


# Driver Code

# Both circle with radius 'r'
# and 'R' have center (0,0)
r = 8; R = 4; r1 = 2; x1 = 6; y1 = 0
if (checkcircle(r, R, r1, x1, y1)):
    print("yes")
else:
    print("no")

# This code is contributed by Smitha Dinesh Semwal.

C#

// C# code to check if a
// circle lies in the ring
using System;

class ring {

    // Function to check if circle
    // lies in the ring
    public static bool checkcircle(int r, int R,
                         int r1, int x1, int y1)
    {
        // distance between center of circle
        // center of concentric circles(origin)
        // using Pythagoras theorem
        int dis = (int)Math.Sqrt(x1 * x1 + y1 * y1);

        // Condition to check if circle
        // is strictly inside the ring
        return (dis - r1 >= R && dis + r1 <= r);
    }

    // Driver Code
    public static void Main()
    {
        // Both circle with radius 'r'
        // and 'R' have center (0, 0)
        int r = 8, R = 4, r1 = 2, x1 = 6, y1 = 0;

        if (checkcircle(r, R, r1, x1, y1))
            Console.WriteLine("yes");
        else
            Console.WriteLine("no");
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP code to check if a circle
// lies in the ring

// Function to check if circle
// lies in the ring
function checkcircle($r, $R, $r1,
                         $x1, $y1)
{

    // distance between center of circle
    // center of concentric circles(origin)
    // using Pythagoras theorem
    $dis = sqrt($x1 * $x1 + $y1 * $y1);

    // Condition to check if circle is
    // strictly inside the ring
    return ($dis-$r1 >= $R && $dis + $r1 <= $r);
}

    // Driver Code
    // Both circle with radius 'r'
    // and 'R' have center (0,0)
    $r = 8; $R = 4;
    $r1 = 2; $x1 = 6;
    $y1 = 0;
    if (checkcircle($r, $R, $r1, $x1, $y1))

    echo "yes" ,"\n";
    else
    echo "no" ,"\n";

// This code is contributed by ajit.
?>

Javascript

<script>
// JavaScript program code to check if a
// circle lies in the ring

// Function to check if circle
    // lies in the ring
    function checkcircle(r, R, r1, x1, y1)
    {
        // distance between center of circle
        // center of concentric circles(origin)
        // using Pythagoras theorem
        let dis = Math.sqrt(x1 * x1 +
                                 y1 * y1);

         // Condition to check if circle
         // is strictly inside the ring
        return (dis - r1 >= R && dis + r1 <= r);
    }

// Driver Code

        // Both circle with radius 'r'
        // and 'R' have center (0,0)
        let r = 8, R = 4, r1 = 2, x1 = 6, y1 = 0;

        if (checkcircle(r, R, r1, x1, y1))
            document.write("yes");
        else
            document.write("no");

// This code is contributed by splevel62.
</script>

Output:

yes

Time Complexity: O(log(n)) since using inbuilt sqrt function

Auxiliary Space: O(1)