寻找关于圆的坐标的象限

给定圆心的半径和坐标。如果点位于圆内，则求另一个给定坐标 (X, Y) 相对于圆心所在的象限。否则打印错误“位于圆圈外”。
如果该点位于圆心输出 0 或如果该点位于任何轴上并且在圆内输出逆时钟方向的下一个象限。

例子：

Input : Centre = (0, 0), Radius = 10 
(X, Y) = (10, 10) 
Output : Lies Outside the Circle 
 

Input : Centre = (0, 3), Radius = 2 
(X, Y) = (1, 4) 
Output : 1 (I quadrant) 
 

方法：
设中心为 (x’, y’)
圆的方程是$(x-x')^2 + (y-y')^2 - r^2 = 0$ –（等式 1）
根据这个方程，
如果$(x-x')^2 + (y-y')^2 > r$点 (x, y) 位于圆外
如果$(x-x')^2 + (y-y')^2 = 0$点 (x, y) 位于圆上
如果$(x-x')^2 + (y-y')^2 < r$点 (x, y) 位于圆内
要检查点相对于圆的位置：-

1. Put given coordinates in equation 1.
2. If it is greater than 0 coordinate lies outside circle.
3. If point lies inside circle find the quadrant within the circle. Check the point 
   with respect to centre of circle. 

以下是上述想法的实现：

C++

// CPP Program to find the quadrant of
// a given coordinate with respect to the
// centre of a circle
#include 
 
using namespace std;
 
// Thus function returns the quadrant number
int getQuadrant(int X, int Y, int R, int PX, int PY)
{
    // Coincides with center
    if (PX == X && PY == Y)
        return 0;
 
    int val = pow((PX - X), 2) + pow((PY - Y), 2);
 
    // Outside circle
    if (val > pow(R, 2))
        return -1;
 
    // 1st quadrant
    if (PX > X && PY >= Y)
        return 1;
 
    // 2nd quadrant
    if (PX <= X && PY > Y)
        return 2;
 
    // 3rd quadrant
    if (PX < X && PY <= Y)
        return 3;
 
    // 4th quadrant
    if (PX >= X && PY < Y)
        return 4;
}
 
// Driver Code
int main()
{
    // Coordinates of centre
    int X = 0, Y = 3;
 
    // Radius of circle
    int R = 2;
 
    // Coordinates of the given point
    int PX = 1, PY = 4;
 
    int ans = getQuadrant(X, Y, R, PX, PY);
    if (ans == -1)
        cout << "Lies Outside the circle" << endl;
    else if (ans == 0)
        cout << "Coincides with centre" << endl;
    else
        cout << ans << " Quadrant" << endl;
    return 0;
}

Java

// Java Program to find the quadrant of
// a given coordinate with respect to the
// centre of a circle
import java.io.*;
class GFG {
 
// Thus function returns
// the quadrant number
static int getQuadrant(int X, int Y,
                       int R, int PX,
                       int PY)
{
     
    // Coincides with center
    if (PX == X && PY == Y)
        return 0;
 
    int val = (int)Math.pow((PX - X), 2) +
              (int)Math.pow((PY - Y), 2);
 
    // Outside circle
    if (val > Math.pow(R, 2))
        return -1;
 
    // 1st quadrant
    if (PX > X && PY >= Y)
        return 1;
 
    // 2nd quadrant
    if (PX <= X && PY > Y)
        return 2;
 
    // 3rd quadrant
    if (PX < X && PY <= Y)
        return 3;
 
    // 4th quadrant
    if (PX >= X && PY < Y)
        return 4;
        return 0;
}
 
    // Driver Code
    public static void main (String[] args)
    {
         
        // Coordinates of centre
        int X = 0, Y = 3;
     
        // Radius of circle
        int R = 2;
     
        // Coordinates of the given point
        int PX = 1, PY = 4;
     
        int ans = getQuadrant(X, Y, R, PX, PY);
        if (ans == -1)
            System.out.println( "Lies Outside the circle");
        else if (ans == 0)
            System.out.println( "Coincides with centre");
        else
            System.out.println( ans +" Quadrant");
    }
}
 
// This code is contributed by anuj_67.

Python3

# Python3 Program to find the
# quadrant of a given coordinate
# w.rt. the centre of a circle
import math
 
# Thus function returns the
# quadrant number
def getQuadrant(X, Y, R, PX, PY):
     
    # Coincides with center
    if (PX == X and PY == Y):
        return 0;
 
    val = (math.pow((PX - X), 2) +
           math.pow((PY - Y), 2));
 
    # Outside circle
    if (val > pow(R, 2)):
        return -1;
 
    # 1st quadrant
    if (PX > X and PY >= Y):
        return 1;
 
    # 2nd quadrant
    if (PX <= X and PY > Y):
        return 2;
 
    # 3rd quadrant
    if (PX < X and PY <= Y):
        return 3;
 
    # 4th quadrant
    if (PX >= X and PY < Y):
        return 4;
 
# Driver Code
# Coordinates of centre
X = 0;
Y = 3;
 
# Radius of circle
R = 2;
 
# Coordinates of the given po
PX = 1;
PY = 4;
 
ans = getQuadrant(X, Y, R, PX, PY);
if (ans == -1) : print("Lies Outside the circle");
elif (ans == 0) : print("Coincides with centre");
else:print(ans, "Quadrant");
 
# This code is contributed by mits

C#

// C# Program to find the quadrant of
// a given coordinate with respect to
// the centre of a circle
using System;
 
class GFG {
 
    // Thus function returns
    // the quadrant number
    static int getQuadrant(int X, int Y,
                  int R, int PX, int PY)
    {
         
        // Coincides with center
        if (PX == X && PY == Y)
            return 0;
     
        int val = (int)Math.Pow((PX - X), 2)
               + (int)Math.Pow((PY - Y), 2);
     
        // Outside circle
        if (val > Math.Pow(R, 2))
            return -1;
     
        // 1st quadrant
        if (PX > X && PY >= Y)
            return 1;
     
        // 2nd quadrant
        if (PX <= X && PY > Y)
            return 2;
     
        // 3rd quadrant
        if (PX < X && PY <= Y)
            return 3;
     
        // 4th quadrant
        if (PX >= X && PY < Y)
            return 4;
            return 0;
    }
 
    // Driver Code
    public static void Main ()
    {
     
        // Coordinates of centre
        int X = 0, Y = 3;
     
        // Radius of circle
        int R = 2;
     
        // Coordinates of the given point
        int PX = 1, PY = 4;
     
        int ans =
             getQuadrant(X, Y, R, PX, PY);
        if (ans == -1)
            Console.WriteLine( "Lies Outside"
                            + " the circle");
        else if (ans == 0)
            Console.WriteLine( "Coincides "
                          + "with centre");
        else
            Console.WriteLine( ans +
                               " Quadrant");
    }
}
 
// This code is contributed by anuj_67.

PHP

 pow($R, 2))
        return -1;
 
    // 1st quadrant
    if ($PX > $X and $PY >= $Y)
        return 1;
 
    // 2nd quadrant
    if ($PX <= $X and $PY > $Y)
        return 2;
 
    // 3rd quadrant
    if ($PX < $X and $PY <= $Y)
        return 3;
 
    // 4th quadrant
    if ($PX >= $X and $PY < $Y)
        return 4;
}
 
    // Driver Code
    // Coordinates of centre
    $X = 0; $Y = 3;
 
    // Radius of circle
    $R = 2;
 
    // Coordinates of the given po$
    $PX = 1;
    $PY = 4;
 
    $ans = getQuadrant($X, $Y, $R,
                         $PX, $PY);
    if ($ans == -1)
        echo "Lies Outside the circle" ;
    else if ($ans == 0)
        echo "Coincides with centre" ;
    else
        echo $ans , " Quadrant" ;
 
// This code is contributed by anuj_67.
?>

Javascript

 
// Javascript Program to find the quadrant of
// a given coordinate with respect to the
// centre of a circle
 
 
// Thus function returns the quadrant number
function getQuadrant( X, Y, R, PX, PY)
{
    // Coincides with center
    if (PX == X && PY == Y)
        return 0;
 
    let val = Math.pow((PX - X), 2) + Math.pow((PY - Y), 2);
 
    // Outside circle
    if (val > Math.pow(R, 2))
        return -1;
 
    // 1st quadrant
    if (PX > X && PY >= Y)
        return 1;
 
    // 2nd quadrant
    if (PX <= X && PY > Y)
        return 2;
 
    // 3rd quadrant
    if (PX < X && PY <= Y)
        return 3;
 
    // 4th quadrant
    if (PX >= X && PY < Y)
        return 4;
}
 
// Driver Code
 
// Coordinates of centre
let X = 0, Y = 3;
 
// Radius of circle
let R = 2;
 
// Coordinates of the given point
let PX = 1, PY = 4;
 
let ans = getQuadrant(X, Y, R, PX, PY);
if (ans == -1)
    document.write("Lies Outside the circle" + "</br>");
else if (ans == 0)
    document.write("Coincides with centre" + "</br>");
else
    document.write(ans + " Quadrant" + "</br>");
 
 

输出：

1 Quadrant