给出两个半径和中心的圆。任务是检查较小的圆圈是否在较大的圆圈内。
例子:
Input: x1 = 10, y1 = 8, x2 = 1, y2 = 2, r1 = 30, r2 = 10
Output: The smaller circle lies completely inside
the bigger circle without touching each other
at a point of circumference.
Input :x1 = 7, y1 = 8;x2 = 3, y2 = 5;r1 = 30, r2 = 25
Output :The smaller circle lies completely inside
the bigger circle with touching each other
at a point of circumference.
方法:
这里可以出现三种情况
- 较小的圆圈完全位于较大的圆圈内,而在圆周点处不会相互接触。
如果发生这种情况,则中心与较小半径之间的距离之和小于较大半径,那么显然较小的圆完全位于圆内,而不接触圆周。
- 较小的圆完全位于较大的圆内部,并且在圆周的某个点相互接触。如果发生这种情况,则中心与较小半径之间的距离之和等于较大半径,那么显然,较小的圆完全位于圆内,并与圆周接触。
- 较小的圆并不完全位于较大的圆内,如果发生这种情况,则中心与较小半径之间的距离之和大于较大圆的半径,则显然较小的圆并不完全位于圆内。
下面是上述方法的实现:
CPP
// C++ program to check if one circle
// lies inside another circle or not.
#include
using namespace std;
void circle(int x1, int y1, int x2,
int y2, int r1, int r2)
{
int distSq = sqrt(((x1 - x2)
* (x1 - x2))
+ ((y1 - y2)
* (y1 - y2)));
if (distSq + r2 == r1)
cout << "The smaller circle lies completely"
<< " inside the bigger circle with "
<< "touching each other "
<< "at a point of circumference. "
<< endl;
else if (distSq + r2 < r1)
cout << "The smaller circle lies completely"
<< " inside the bigger circle without"
<< " touching each other "
<< "at a point of circumference. "
<< endl;
else
cout << "The smaller does not lies inside"
<< " the bigger circle completely."
<< endl;
}
// Driver code
int main()
{
int x1 = 10, y1 = 8;
int x2 = 1, y2 = 2;
int r1 = 30, r2 = 10;
circle(x1, y1, x2, y2, r1, r2);
return 0;
}
Java
// Java program to check if one circle
// lies inside another circle or not.
import java.io.*;
class GFG
{
static void circle(int x1, int y1, int x2,
int y2, int r1, int r2)
{
int distSq = (int)Math.sqrt(((x1 - x2)
* (x1 - x2))
+ ((y1 - y2)
* (y1 - y2)));
if (distSq + r2 == r1)
{
System.out.println("The smaller circle lies completely"
+ " inside the bigger circle with "
+ "touching each other "
+ "at a point of circumference. ") ;
}
else if (distSq + r2 < r1)
{
System.out.println("The smaller circle lies completely"
+ " inside the bigger circle without"
+ " touching each other "
+ "at a point of circumference.") ;
}
else
{
System.out.println("The smaller does not lies inside"
+ " the bigger circle completely.") ;
}
}
// Driver code
public static void main (String[] args)
{
int x1 = 10, y1 = 8;
int x2 = 1, y2 = 2;
int r1 = 30, r2 = 10;
circle(x1, y1, x2, y2, r1, r2);
}
}
// This code is contributed by ajit_00023.
Python
# Python3 program to check if one circle
# lies inside another circle or not.
def circle(x1, y1, x2,y2, r1, r2):
distSq = (((x1 - x2)* (x1 - x2))+ ((y1 - y2)* (y1 - y2)))**(.5)
if (distSq + r2 == r1):
print("The smaller circle lies completely"
" inside the bigger circle with "
"touching each other "
"at a poof circumference. ")
elif (distSq + r2 < r1):
print("The smaller circle lies completely"
" inside the bigger circle without"
" touching each other "
"at a poof circumference. ")
else:
print("The smaller does not lies inside"
" the bigger circle completely.")
# Driver code
x1 ,y1 = 10,8
x2 ,y2 = 1, 2
r1 ,r2 = 30,10
circle(x1, y1, x2, y2, r1, r2)
# This code is contributed by mohit kumar 29
C#
// C# program to check if one circle
// lies inside another circle or not.
using System;
class GFG
{
static void circle(int x1, int y1, int x2,
int y2, int r1, int r2)
{
int distSq = (int)Math.Sqrt(((x1 - x2)
* (x1 - x2))
+ ((y1 - y2)
* (y1 - y2)));
if (distSq + r2 == r1)
{
Console.WriteLine("The smaller circle lies completely"
+ " inside the bigger circle with "
+ "touching each other "
+ "at a point of circumference. ") ;
}
else if (distSq + r2 < r1)
{
Console.WriteLine("The smaller circle lies completely"
+ " inside the bigger circle without"
+ " touching each other "
+ "at a point of circumference.") ;
}
else
{
Console.WriteLine("The smaller does not lies inside"
+ " the bigger circle completely.") ;
}
}
// Driver code
static public void Main ()
{
int x1 = 10, y1 = 8;
int x2 = 1, y2 = 2;
int r1 = 30, r2 = 10;
circle(x1, y1, x2, y2, r1, r2);
}
}
// This code is contributed by AnkitRai01
PHP
Javascript
输出:
较小的圆完全位于较大的圆内,而在圆周点上不会相互接触。
较小的圆完全位于较大的圆内,而在圆周点上不会相互接触。