Given two circles with a given radius and centers. The task is to find the number of common tangents between these circles.
Examples:
Input: x1 = -10, y1 = 8, x2 = 14, y2 = -24, r1 = 30, r2 = 10 Output: 3 Input: x1 = 40, y1 = 8, x2 = 14, y2 = 54, r1 = 39, r2 = 51 Output: 2
Approach:
- First of all we will check whether the circles touch each other externally, intersect each other or do not touch each other at all.(Please refer here)
- Then if the circles do not touch each other externally, then obviously they will have 4 common tangents, two direct and two transverse.
- If the circles touch each other externally, then they will have 3 common tangents, two direct and one transverse.
The tangent in between can be thought of as the transverse tangents coinciding together.
- If the circles intersect each other, then they will have 2 common tangents, both of them will be direct.
- If one circle is inside another circle, then they will have only one common tangent
Algorithm:
Step 1: Create the “circle” function, which has six inputs: x1, y1, x2, y2, r1, and r2.
Step 2: Use the following formula to determine the separation between the centers of the two circles: (x1 – x2)^2 + (y1 – y2)^2. Use the formula (r1 + r2)^2 to determine the total of the two circles’ radii.
Step 3: To ascertain the relationship between the two circles, compare the computed distance and the sum of radii: a. If the distance is equal to the sum of radii, return 1 as there is only one common tangent.
b. If the distance is greater than the sum of radii, return -1 as there are four common tangents.
c. If the distance is less than the sum of radii, return 0 as there are two common tangents.
Below is the implementation of the above approach:
C++
// C++ program to find// the number of common tangents// between the two circles#include <bits/stdc++.h>using namespace std;int circle(int x1, int y1, int x2, int y2, int r1, int r2){ int distSq = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2); int radSumSq = (r1 + r2) * (r1 + r2); if (distSq == radSumSq) return 1; else if (distSq > radSumSq) return -1; else return 0;}// Driver codeint main(){ int x1 = -10, y1 = 8; int x2 = 14, y2 = -24; int r1 = 30, r2 = 10; int t = circle(x1, y1, x2, y2, r1, r2); if (t == 1) cout << "There are 3 common tangents" << " between the circles."; else if (t < 0) cout << "There are 4 common tangents" << " between the circles."; else cout << "There are 2 common tangents" << " between the circles."; return 0;} |
Java
// Java program to find// the number of common tangents// between the two circlesimport java.io.*;class GFG{ static int circle(int x1, int y1, int x2, int y2, int r1, int r2){ int distSq = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2); int radSumSq = (r1 + r2) * (r1 + r2); if (distSq == radSumSq) return 1; else if (distSq > radSumSq) return -1; else return 0;}// Driver codepublic static void main (String[] args){ int x1 = -10, y1 = 8; int x2 = 14, y2 = -24; int r1 = 30, r2 = 10; int t = circle(x1, y1, x2, y2, r1, r2); if (t == 1) System.out.println ("There are 3 common tangents"+ " between the circles."); else if (t < 0) System.out.println ("There are 4 common tangents"+ " between the circles."); else System.out.println ("There are 2 common tangents" + " between the circles."); }}// This code is contributed by ajit. |
Python3
# Python3 program to find# the number of common tangents# between the two circlesdef circle(x1, y1, x2,y2, r1, r2): distSq = (x1 - x2) * (x1 - x2)+ (y1 - y2) * (y1 - y2) radSumSq = (r1 + r2) * (r1 + r2) if (distSq == radSumSq): return 1 elif (distSq > radSumSq): return -1 else: return 0# Driver codex1,y1 = -10,8;x2,y2 = 14,-24;r1,r2 = 30,10;t = circle(x1, y1, x2,y2, r1, r2);if (t == 1): print("There are 3 common tangents between the circles.")elif (t < 0): print("There are 4 common tangents between the circles.")else: print("There are 2 common tangents between the circles.")# This code is contributed by mohit kumar 29 |
C#
// C# program to find// the number of common tangents// between the two circlesusing System;class GFG{ static int circle(int x1, int y1, int x2, int y2, int r1, int r2){ int distSq = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2); int radSumSq = (r1 + r2) * (r1 + r2); if (distSq == radSumSq) return 1; else if (distSq > radSumSq) return -1; else return 0;}// Driver codepublic static void Main (String []args){ int x1 = -10, y1 = 8; int x2 = 14, y2 = -24; int r1 = 30, r2 = 10; int t = circle(x1, y1, x2, y2, r1, r2); if (t == 1) Console.WriteLine ("There are 3 common tangents"+ " between the circles."); else if (t < 0) Console.WriteLine ("There are 4 common tangents"+ " between the circles."); else Console.WriteLine ("There are 2 common tangents" + " between the circles.");}}// This code is contributed by Arnab Kundu |
PHP
<?php// PHP program to find// the number of common tangents// between the two circlesfunction circle($x1, $y1, $x2, $y2, $r1, $r2){ $distSq = ($x1 - $x2) * ($x1 - $x2) + ($y1 - $y2) * ($y1 - $y2); $radSumSq = ($r1 + $r2) * ($r1 + $r2); if ($distSq == $radSumSq) return 1; else if ($distSq > $radSumSq) return -1; else return 0;} // Driver code $x1 = -10; $y1 = 8; $x2 = 14; $y2 = -24; $r1 = 30; $r2 = 10; $t = circle($x1, $y1, $x2, $y2, $r1, $r2); if ($t == 1) echo "There are 3 common tangents" ," between the circles."; else if ($t < 0) echo "There are 4 common tangents" , " between the circles."; else echo "There are 2 common tangents" , " between the circles."; // This code is contributed by AnkitRai01?> |
Javascript
<script>// Javascript program to find// the number of common tangents// between the two circlesfunction circle(x1, y1, x2, y2, r1, r2){ var distSq = (x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2); var radSumSq = (r1 + r2) * (r1 + r2); if (distSq == radSumSq) return 1; else if (distSq > radSumSq) return -1; else return 0;}// Driver code var x1 = -10, y1 = 8; var x2 = 14, y2 = -24; var r1 = 30, r2 = 10; var t = circle(x1, y1, x2, y2, r1, r2); if (t == 1) document.write("There are 3 common tangents between the circles."); else if (t < 0) document.write("There are 4 common tangents between the circles."); else document.write("There are 2 common tangents between the circles.");</script> |
Output:
There are 3 common tangents between the circles.
Time Complexity: O(1)
Auxiliary Space: O(1)
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