SOLUTION: Hi, 3x^2 + 3y^2 - 6x - 6 = 0 (describe and sketch the given equation) I believe this is a circle and I've tried "completing the square", but without the b(y) part of the equa

Algebra ->  Graphs -> SOLUTION: Hi, 3x^2 + 3y^2 - 6x - 6 = 0 (describe and sketch the given equation) I believe this is a circle and I've tried "completing the square", but without the b(y) part of the equa      Log On


   

Question 187674This question is from textbook Precalculus
: Hi,
3x^2 + 3y^2 - 6x - 6 = 0 (describe and sketch the given equation)
I believe this is a circle and I've tried "completing the square", but without the b(y) part of the equation, I'm lost. Please help.
Thank you,
Lynn This question is from textbook Precalculus

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
3x^2 + 3y^2 - 6x - 6 = 0 (describe and sketch the given equation)
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Divide thru by 3 to get:
x^2 + y^2 - 2x - 2 = 0
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x^2 - 2x + ? + y^2 = 2
x^2 - 2x + 1 + y^2 = 3
(x-1)^2 + (y-0)^2 = 3
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Circle with center at (1,0) and radius = sqrt(3)
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Cheers,
Stan H.