给定圆心和半径的两个圆之间的公共切线数
原文:https://www . geesforgeks . org/两个圆之间的公共切线数-如果给定它们的中心和半径/
给定两个具有给定半径和中心的圆。任务是找到这些圆之间的公共切线的数量。 例:
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
接近 :
- 首先我们会检查圆是否在外部相互接触，相互相交或者根本不接触。(请参考此处)
- 那么如果圆在外部没有相互接触，那么显然它们将有 4 条公共切线，两条直切线和两条横切线。
- 如果圆在外部相互接触，那么它们将有 3 条公共切线，两条直切线和一条横切线。 中间的切线可以认为是重合在一起的横向切线。
- 如果圆彼此相交，那么它们将有 2 条公共切线，两者都是直的。
- 如果一个圆在另一个圆内，那么它们只有一个公共切线。
下面是上述方法的实现:
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 语言(一种计算机语言，尤用于创建网站)
// 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.
Python 3
# 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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?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
?>
java 描述语言
<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>
输出:
There are 3 common tangents between the circles.
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