Given the center of circle (x1, y1) and its radius r, find the equation of the circle having center (x1, y1) and having radius r.
Examples:
Input : x1 = 2, y1 = -3, r = 8
Output : x^2 + y^2 – 4*x + 6*y = 51.
Input : x1 = 0, y1 = 0, r = 2
Output : x^2 + y^2 – 0*x + 0*y = 4.
Approach:
Given the center of circle (x1, y1) and its radius r, we have to find the equation of the circle having center (x1, y1) and having radius r.
the equation of circle having center (x1, y1) and having radius r is given by :-
on expanding above equation
on arranging above we get
Below is the implementation of above approach:
C++
// CPP program to find the equation// of circle.#include <iostream>using namespace std;// Function to find the equation of circlevoid circle_equation(double x1, double y1, double r){ double a = -2 * x1; double b = -2 * y1; double c = (r * r) - (x1 * x1) - (y1 * y1); // Printing result cout << "x^2 + (" << a << " x) + "; cout << "y^2 + (" << b << " y) = "; cout << c << "." << endl;}// Driver codeint main(){ double x1 = 2, y1 = -3, r = 8; circle_equation(x1, y1, r); return 0;} |
Java
// Java program to find the equation// of circle.import java.util.*;class solution{ // Function to find the equation of circlestatic void circle_equation(double x1, double y1, double r){ double a = -2 * x1; double b = -2 * y1; double c = (r * r) - (x1 * x1) - (y1 * y1); // Printing result System.out.print("x^2 + (" +a+ " x) + "); System.out.print("y^2 + ("+b + " y) = "); System.out.println(c +"." );}// Driver codepublic static void main(String arr[]){ double x1 = 2, y1 = -3, r = 8; circle_equation(x1, y1, r); }} |
Python3
# Python3 program to find the # equation of circle.# Function to find the # equation of circledef circle_equation(x1, y1, r): a = -2 * x1; b = -2 * y1; c = (r * r) - (x1 * x1) - (y1 * y1); # Printing result print("x^2 + (", a, "x) + ", end = ""); print("y^2 + (", b, "y) = ", end = ""); print(c, ".");# Driver codex1 = 2; y1 = -3; r = 8;circle_equation(x1, y1, r);# This code is contributed # by mits |
C#
// C# program to find the equation// of circle.using System;class GFG{// Function to find the equation of circlepublic static void circle_equation(double x1, double y1, double r){ double a = -2 * x1; double b = -2 * y1; double c = (r * r) - (x1 * x1) - (y1 * y1); // Printing result Console.Write("x^2 + (" + a + " x) + "); Console.Write("y^2 + ("+ b + " y) = "); Console.WriteLine(c + "." );}// Driver codepublic static void Main(string []arr){ double x1 = 2, y1 = -3, r = 8; circle_equation(x1, y1, r);}}// This code is contributed // by SoumkMondal |
PHP
<?php// PHP program to find the equation// of circle.// Function to find the // equation of circlefunction circle_equation($x1, $y1, $r){ $a = -2 * $x1; $b = -2 * $y1; $c = ($r * $r) - ($x1 * $x1) - ($y1 * $y1); // Printing result echo "x^2 + (" . $a . " x) + "; echo "y^2 + (" . $b . " y) = "; echo $c . "." . "\n";}// Driver code$x1 = 2; $y1 = -3; $r = 8;circle_equation($x1, $y1, $r);// This code is contributed // by Akanksha Rai?> |
Javascript
<script>// java script program to find the equation// of circle.// Function to find the// equation of circlefunction circle_equation(x1, y1, r){ let a = -2 * x1; let b = -2 * y1; let c = (r * r) - (x1 * x1) - (y1 * y1); // Printing result document.write( "x^2 + (" +a + " x) + "); document.write( "y^2 + (" +b+ " y) = "); document.write( c+ "<br>");}// Driver codelet x1 = 2; let y1 = -3; let r = 8;circle_equation(x1, y1, r);// This code is contributed by sravan kumar</script> |
Output:
x^2 + (-4 x) + y^2 + (6 y) = 51.
Time complexity: O(1), since there is no loop or recursion.
Auxiliary Space: O(1), since no extra space has been taken.
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