Question 1090252
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<pre>
{{{x^2 + 4x + y^2 - 4y}}} = 4,    (1)
{{{(x-2)^2 + (y-2)^2}}} = 4.    (2)

Open the parentheses in equation (2). You will get

{{{x^2 + 4x + y^2 - 4y}}} = 4,    (1')
{{{x^2 - 4x + y^2 - 4y}}} = -4.   (2')


Subtract eq(1') from eq(2') (both sides). You will get

-8x = -4 - 4 = -8.


Hence, x = 1.


Now, substitute x= 1 into either of the two equations (1') or (2') and solve for y.


Can you complete these simple calculations on your own ?
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For many similar solved problems/samples see the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Systems-of-equations/Solving-the-system-of-algebraic-equations-of-degree-2.lesson>Solving the system of algebraic equations of degree 2</A>, 

in this site.



Also, you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lesson is the part of this online textbook under the topic "<U>Systems of equations that are not linear</U>".