Given Centre coordinate (c1, c2) and one coordinate (x1, y1) of the diameter of a circle, find the other end coordinate point (x2, y2) of diameter.
Examples:
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
The Midpoint Formula:
The midpoint of two ends coordinates points, (x1, y2) and (x2, y2) is the point M can be found by using:
We have need of a (x2, y2) coordinates, so we apply the midpoint the formula
c1 = ((x1+x2)/2), c2 = ((y1+y2)/2)
2*c1 = (x1+x2), 2*c2 = (y1+y2)
x2 = (2*c1 - x1), y2 = (2*c2 - y1)
C++
// CPP program to find the // other-end point of diameter#include <iostream>using namespace std;// function to find the// other-end point of diametervoid 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 Programint main(){ int x1 = -4, y1 = -1; int c1 = 3, c2 = 5; endPointOfDiameterofCircle(x1, y1, c1, c2); return 0;} |
Java
// Java program to find the other-end point of // diameterimport 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. |
Python3
# Python3 program to find the # other-end point of diameter# function to find the# other-end point of diameterdef 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 Programx1 = -4y1 = -1c1 = 3c2 = 5 endPointOfDiameterofCircle(x1, y1, c1, c2) # This code is contributed by Smitha. |
C#
// C# program to find the other - // end point of diameterusing 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. |
PHP
<?php// PHP program to find the // other-end point of diameter// function to find the// other-end point of diameterfunction 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?> |
Javascript
<script>// Javascript program to find the // other-end point of diameter// Function to find the// other-end point of diameterfunction 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> |
Output
x2 = 10 y2 = 11
Time complexity: O(1)
Auxiliary Space: O(1)
Similarly if we given a centre (c1, c2) and other end coordinate (x2, y2) of a diameter and we finding a (x1, y1) coordinates
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)
So The other end coordinates (x1, y1) of a diameter is
x1 = (2*c1 - x2), y1 = (2*c2 - y2)
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