求圆内直径的另一端坐标
给定圆心坐标(c1，c2)和圆直径的一个坐标(x1，y1)，求直径的另一个端点坐标点(x2，y2)。
示例:
Input : x1 = –1, y1 = 2, and c1 = 3, c2 = –6
Output : x2 = 7, y2 = -14
Input : x1 = 6.4, y1 = 3 and c1 = –10.7, c2 = 4
Output : x2 = -27.8, y2 = 5
中点公式:
两端坐标点的中点，(x1，y2)和(x2，y2)是 M 点可以用:
找到
我们需要一个(x2，y2)坐标，所以我们应用公式的中点
c1 = ((x1+x2)/2), c2 = ((y1+y2)/2)
2*c1 = (x1+x2), 2*c2 = (y1+y2)
x<sub>2 = (2*c1 - x1), y2 = (2*c2 - y1)</sub>
C++
// CPP program to find the
// other-end point of diameter
#include <iostream>
using namespace std;
// function to find the
// other-end point of diameter
void endPointOfDiameterofCircle(int x1,
int y1, int c1, int c2)
{
// find end point for x coordinates
cout << "x2 = "
<< (float)(2 * c1 - x1)<< " ";
// find end point for y coordinates
cout << "y2 = " << (float)(2 * c2 - y1);
}
// Driven Program
int main()
{
int x1 = -4, y1 = -1;
int c1 = 3, c2 = 5;
endPointOfDiameterofCircle(x1, y1, c1, c2);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find the other-end point of
// diameter
import java.io.*;
class GFG {
// function to find the other-end point of
// diameter
static void endPointOfDiameterofCircle(int x1,
int y1, int c1, int c2)
{
// find end point for x coordinates
System.out.print( "x2 = "
+ (2 * c1 - x1) + " ");
// find end point for y coordinates
System.out.print("y2 = " + (2 * c2 - y1));
}
// Driven Program
public static void main (String[] args)
{
int x1 = -4, y1 = -1;
int c1 = 3, c2 = 5;
endPointOfDiameterofCircle(x1, y1, c1, c2);
}
}
// This code is contributed by anuj_67.
Python 3
# Python3 program to find the
# other-end point of diameter
# function to find the
# other-end point of diameter
def endPointOfDiameterofCircle(x1, y1, c1, c2):
# find end point for x coordinates
print("x2 =", (2 * c1 - x1), end=" ")
# find end point for y coordinates
print("y2 =" , (2 * c2 - y1))
# Driven Program
x1 = -4
y1 = -1
c1 = 3
c2 = 5
endPointOfDiameterofCircle(x1, y1, c1, c2)
# This code is contributed by Smitha.
C
// C# program to find the other -
// end point of diameter
using System;
class GFG {
// function to find the other - end
// point of diameter
static void endPointOfDiameterofCircle(int x1,
int y1,
int c1,
int c2)
{
// find end point for x coordinates
Console.Write("x2 = "+ (2 * c1 - x1) + " ");
// find end point for y coordinates
Console.Write("y2 = " + (2 * c2 - y1));
}
// Driver Code
public static void Main ()
{
int x1 = -4, y1 = -1;
int c1 = 3, c2 = 5;
endPointOfDiameterofCircle(x1, y1, c1, c2);
}
}
// This code is contributed by anuj_67.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find the
// other-end point of diameter
// function to find the
// other-end point of diameter
function endPointOfDiameterofCircle($x1,
$y1, $c1, $c2)
{
// find end point for x coordinates
echo "x2 = ",(2 * $c1 - $x1)," ";
// find end point for y coordinates
echo "y2 = " , (2 * $c2 - $y1);
}
// Driven Program
$x1 = -4;
$y1 = -1;
$c1 = 3;
$c2 = 5;
endPointOfDiameterofCircle($x1, $y1,
$c1, $c2);
// This code is contributed by Smitha
?>
java 描述语言
<script>
// Javascript program to find the
// other-end point of diameter
// Function to find the
// other-end point of diameter
function endPointOfDiameterofCircle(x1, y1, c1, c2)
{
// Find end point for x coordinates
document.write("x2 = " + (2 * c1 - x1) + " ");
// Find end point for y coordinates
document.write("y2 = " + (2 * c2 - y1));
}
// Driver code
let x1 = -4, y1 = -1;
let c1 = 3, c2 = 5;
endPointOfDiameterofCircle(x1, y1, c1, c2);
// This code is contributed by jana_sayantan
</script>
输出
x2 = 10 y2 = 11
类似地，如果我们给定一个直径的中心(c1，c2)和另一端坐标(x2，y2)，我们找到一个(x1，y1)坐标
Proof for (x1, y1) :
c1 = ((x1+x2)/2), c2 = ((y1+y2)/2)
2*c1 = (x1+x2), 2*c2 = (y1+y2)
x1 = (2*c1 - x2), y1 = (2*c2 - y2)
所以直径的另一端坐标(x1，y1)是
x1 = (2*c1 - x2), y1 = (2*c2 - y2)
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