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stupidinmath Group Title

If the center of the circle x^2 + y^2 + ax + by + 2 = 0 is point (4,-8), what is a + b? a. -8 b. -4 c. 4 d. 8 e. 24

  • 7 months ago
  • 7 months ago

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  1. MrNood Group Title
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    It is rare in life to find someone who would call themselves 'stupid in English' or 'stupid in sport' Maths is only another skill to be learned and enjoyed like any other learning. It is not a badge of pride to 'not be able to do maths'

    • 7 months ago
  2. stupidinmath Group Title
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    i know..

    • 7 months ago
  3. surjithayer Group Title
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    complete the squares and equate the centre.

    • 7 months ago
  4. mathmale Group Title
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    I'd go so far as to try opening a new account with OpenStudy, so that you don't go on labeling yourself with such a derogatory term. It's not funny and does you no good.

    • 7 months ago
  5. mathmale Group Title
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    Hint: the equation of a circle with radius r and centered at (h,k) is\[(x-h)^{2}+(y-k)^{2}=r ^{2}\]. Hope this puts you on track towards finding a solution to this problem.

    • 7 months ago
  6. mathmale Group Title
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    Please demo how you'd complete the square: x^2 + 6x

    • 7 months ago
  7. RadEn Group Title
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    just an alternative : if given the equation of circle x^2 + y^2 + ax + by + c = 0, then the centre is (-a/2 , -b/2). knowed the centre is (4, - 8) so, -a/2 = 4 solve for a also -b/2 = -8 solve for b after that, you can determine the value of a + b

    • 7 months ago
  8. stupidinmath Group Title
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    i think its a..?

    • 7 months ago
  9. stupidinmath Group Title
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    -8

    • 7 months ago
  10. RadEn Group Title
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    have you solve them : -a/2 = 4 a = ... -b/2 = -8 b = ...

    • 7 months ago
  11. surjithayer Group Title
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    \[\left( x^2+ax+\left( \frac{ a }{2 } \right)^2 \right)+\left( y^2+by+\left( \frac{ b }{ 2 } \right)^2 \right)=-2+\left( \frac{ a }{ 2 } \right)^2+\left( \frac{ b }{2 } \right)^2\] \[\left( x-\frac{ a }{2 } \right)^2+\left( y-\frac{ b }{ 2 } \right)^2=-2+\frac{ a^2 }{4 }+\frac{ b^2 }{4 }\] centre is \[\left( \frac{ -a }{ 2 } ,\frac{ -b }{2 }\right)\] \[\frac{ -a }{ 2 }=4,a=-8,\frac{ -b }{2 }=-8,b=-8*-2=16\]

    • 7 months ago
  12. stupidinmath Group Title
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    -16 - 8.

    • 7 months ago
  13. stupidinmath Group Title
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    oh... positive 8?

    • 7 months ago
  14. mathmale Group Title
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    Dear Math Whiz: Why not try doing some of these calculations yourself and then sharing the results? Those trying to help you could then give you more specific and useful feedback.

    • 7 months ago
  15. surjithayer Group Title
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    we can also find by comparing with \[x^2+y^2+2gx+2fy+c=0,where~centre~is~\left( -g,-f \right) ~and radius=\sqrt{g^2+f^2-c}\]

    • 7 months ago
  16. surjithayer Group Title
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    radius=\[\sqrt{g^2+f^2-c}\]

    • 7 months ago
  17. mathmale Group Title
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    I still think completing the square is a more readily understandable approach. But first, Math Whiz (I refuse to call you stupidinmath), are you familiar with completing the square?

    • 7 months ago
  18. MrNood Group Title
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    Welcome to Math Whiz

    • 7 months ago
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