Dados dos círculos con un radio y centros dados. La tarea es encontrar el número de tangentes comunes entre estos círculos.
Ejemplos:
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
Enfoque :
- En primer lugar, comprobaremos si los círculos se tocan entre sí externamente, si se cruzan entre sí o si no se tocan en absoluto. ( Consulte aquí )
- Entonces, si los círculos no se tocan externamente, obviamente tendrán 4 tangentes comunes, dos directas y dos transversales.
- Si los círculos se tocan externamente, entonces tendrán 3 tangentes comunes, dos directas y una transversal.
La tangente intermedia se puede considerar como las tangentes transversales que coinciden entre sí.
- Si los círculos se cortan entre sí, entonces tendrán 2 tangentes comunes, ambas serán directas.
- Si un círculo está dentro de otro círculo, solo tendrán una tangente común.
A continuación se muestra la implementación del enfoque anterior:
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 code
int 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 circles
import 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 code
public 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 circles
def 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 code
x1,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 circles
using 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 code
public 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 circles
function 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 circles
function 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>
Producción:
There are 3 common tangents between the circles.
Complejidad de tiempo: O(1)
Espacio Auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por IshwarGupta y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA