How do I write the equation of the circle in standard form, and determine the center and radius if it exists? for: 2x^2 + 2y^2 + 8x - 12y - 8 = 0

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How do I write the equation of the circle in standard form, and determine the center and radius if it exists? for: 2x^2 + 2y^2 + 8x - 12y - 8 = 0

Mathematics
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complete the squares if need be
2(x^2 +4x) + 2(y^2 -6y) = 8 2(x^2 +4x +4)-8 + 2(y^2 -6y+9)-18 = 8
2(x+2)^2 + 2(y-3)^2 = 8+8+18 = 34

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Other answers:

the center is the opposites of the x and y modifiers... C=(-2,3)
r^2 = 34 r = sqrt(34)
or do we divide it all by 2? first?
r=sqrt(17) prolly
Not easy, so do they both get divided by 2 then?
everything gets divided by 2...everything
Okay,
on a side note, we could divide everying by 34 to get: (x+2)^2 (y-3)^2 ------- + ------- = 1 ; which is the same equation 17 17
so I would stick with r = sqrt(17)
Okay, thank you
youre welcome :)

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