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anonymous

  • 2 years ago

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

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  1. MrNood
    • 2 years ago
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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'

  2. anonymous
    • 2 years ago
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    i know..

  3. anonymous
    • 2 years ago
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    complete the squares and equate the centre.

  4. mathmale
    • 2 years ago
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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.

  5. mathmale
    • 2 years ago
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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.

  6. mathmale
    • 2 years ago
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    Please demo how you'd complete the square: x^2 + 6x

  7. RadEn
    • 2 years ago
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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

  8. anonymous
    • 2 years ago
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    i think its a..?

  9. anonymous
    • 2 years ago
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    -8

  10. RadEn
    • 2 years ago
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    have you solve them : -a/2 = 4 a = ... -b/2 = -8 b = ...

  11. anonymous
    • 2 years ago
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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 \[\left( \frac{ -a }{ 2 } ,\frac{ -b }{2 }\right)\] \[\frac{ -a }{ 2 }=4,a=-8,\frac{ -b }{2 }=-8,b=-8*-2=16\]

  12. anonymous
    • 2 years ago
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    -16 - 8.

  13. anonymous
    • 2 years ago
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    oh... positive 8?

  14. mathmale
    • 2 years ago
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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.

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

  16. anonymous
    • 2 years ago
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    radius=\[\sqrt{g^2+f^2-c}\]

  17. mathmale
    • 2 years ago
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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?

  18. MrNood
    • 2 years ago
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    Welcome to Math Whiz

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