Finding Quadrant of a Coordinate with respect to a Circle
Given the radius and coordinates of the Centre of a circle. Find the quadrant in which another given coordinate (X, Y) lies with respect to the Centre of the circle if the point lies inside the circle. Else print an error “Lies outside the circle”.
If the point lies at the Centre of circle output 0 or if the point lies on any of the axes and inside the circle output the next quadrant in anti-clock direction.
Examples:
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)
Approach:
Let center be (x’, y’)
Equation of circle is – (Eq. 1)
According to this equation,
If r " title="Rendered by QuickLaTeX.com" height="30" width="277" style="vertical-align: -7px;">point (x, y) lies outside of circle
If point (x, y) lies on the circle
If point (x, y) lies inside of circle
To check position of point with respect to circle:-
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.
Below is the implementation of above idea :
C++
// CPP Program to find the quadrant of // a given coordinate with respect to the // centre of a circle #include <bits/stdc++.h> using namespace std; // This 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
<?php // PHP 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 and $PY == $Y ) return 0; $val = pow(( $PX - $X ), 2) + 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) 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
<script> // 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>" ); </script> |
Output
1 Quadrant
Time complexity: O(1)
Auxiliary space: O(1)
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