anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
complete the squares if need be
amistre64
  • amistre64
2(x^2 +4x) + 2(y^2 -6y) = 8 2(x^2 +4x +4)-8 + 2(y^2 -6y+9)-18 = 8
amistre64
  • amistre64
2(x+2)^2 + 2(y-3)^2 = 8+8+18 = 34

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amistre64
  • amistre64
the center is the opposites of the x and y modifiers... C=(-2,3)
amistre64
  • amistre64
r^2 = 34 r = sqrt(34)
amistre64
  • amistre64
or do we divide it all by 2? first?
amistre64
  • amistre64
r=sqrt(17) prolly
anonymous
  • anonymous
Not easy, so do they both get divided by 2 then?
amistre64
  • amistre64
everything gets divided by 2...everything
anonymous
  • anonymous
Okay,
amistre64
  • amistre64
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
amistre64
  • amistre64
so I would stick with r = sqrt(17)
anonymous
  • anonymous
Okay, thank you
amistre64
  • amistre64
youre welcome :)

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