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anonymous

  • 5 years ago

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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  1. amistre64
    • 5 years ago
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    complete the squares if need be

  2. amistre64
    • 5 years ago
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    2(x^2 +4x) + 2(y^2 -6y) = 8 2(x^2 +4x +4)-8 + 2(y^2 -6y+9)-18 = 8

  3. amistre64
    • 5 years ago
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    2(x+2)^2 + 2(y-3)^2 = 8+8+18 = 34

  4. amistre64
    • 5 years ago
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    the center is the opposites of the x and y modifiers... C=(-2,3)

  5. amistre64
    • 5 years ago
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    r^2 = 34 r = sqrt(34)

  6. amistre64
    • 5 years ago
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    or do we divide it all by 2? first?

  7. amistre64
    • 5 years ago
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    r=sqrt(17) prolly

  8. anonymous
    • 5 years ago
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    Not easy, so do they both get divided by 2 then?

  9. amistre64
    • 5 years ago
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    everything gets divided by 2...everything

  10. anonymous
    • 5 years ago
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    Okay,

  11. amistre64
    • 5 years ago
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    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

  12. amistre64
    • 5 years ago
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    so I would stick with r = sqrt(17)

  13. anonymous
    • 5 years ago
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    Okay, thank you

  14. amistre64
    • 5 years ago
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    youre welcome :)

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