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Mathematics
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find the center and radius of the circle 2x^2+2y^2-2x+8y=1/2
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lets start by completing the squares. 2y^2 +8y = 2(y^2+4y) = 2(y^2+4y+4)-8 = 2(y+2)^2 -8
2x^2-2x = 2(x^2-x) = 2(x^2-x+1/4) -1/2 = 2(x-1/2)^2 -1/2
so your expression then becomes 2(y+2)^2 +2(x-1/2)^2 -8 =1 (y+2)^2+(x-1/2)^2 = 9/2
so the center is (1/2,-2) and the radius is (3/sqrt2)
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ya thanks
great!
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