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Find the center of the circle defined by the equation x^2+y^2-4x-4y+4=0
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You want to complete the square for x and y. \[x^2-4x+(4-4)+y^2-4y+(4-4)+4=(x-2)^2+(y-2)^2=4\]
So your center is (2,2)
^ was going to write the exact thing lol
:P
Thanks love!
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No problem :P
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