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iheartfood

What is the value of a for the attached circle in general form?

  • one year ago
  • one year ago

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  1. iheartfood
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    • one year ago
  2. iheartfood
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    **no answer choices!! and idk how to solve :(

    • one year ago
  3. UnkleRhaukus
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    factor the equation using prefect squares

    • one year ago
  4. iheartfood
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    which equation??

    • one year ago
  5. iheartfood
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    like the one of the circle?? (which i find??) or the general form?

    • one year ago
  6. UnkleRhaukus
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    \[x^2+ax+y^2+by+c=0\] \[\left(x+\frac a2\right)^2-a^2+y^2+by+c=0\]

    • one year ago
  7. iheartfood
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    why whoops?? haha :P am i not seeing something ur seeing??

    • one year ago
  8. UnkleRhaukus
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    (i forgot to square ) (fixed now)

    • one year ago
  9. iheartfood
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    ohhh okay i see :P so what do i do now?

    • one year ago
  10. UnkleRhaukus
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    factor the y terms using the same technique , then move the three constants outside the brackets to the right hand side

    • one year ago
  11. iheartfood
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    (x+a/x)^2=(-a+y+by)^2 ?? did i do that right?? :/ idk...

    • one year ago
  12. iheartfood
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    oops i forgot the +c... so like this?? (x+a/x)^2=(-a+y+by)^2+c ??? idk :(

    • one year ago
  13. UnkleRhaukus
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    \[\left(x+\frac a2\right)^2-a^2+y^2+by+c=0\] \[\left(x+\frac a2\right)^2-a^2+\left(y+\frac b2\right)^2-b^2+c=0\] \[\left(x+\frac a2\right)^2+\left(y+\frac b2\right)^2=a^2+b^2-c\]

    • one year ago
  14. iheartfood
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    oh okay :P but how do i solve for a?

    • one year ago
  15. UnkleRhaukus
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    compair to the general form of a circle

    • one year ago
  16. iheartfood
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    |dw:1354600365624:dw|

    • one year ago
  17. iheartfood
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    or the one in my question?

    • one year ago
  18. UnkleRhaukus
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    |dw:1354600399190:dw|

    • one year ago
  19. iheartfood
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    oh sorry! i meant - haha :P so do i refer to that one?? or the one in my problem? this one?|dw:1354600447714:dw|

    • one year ago
  20. UnkleRhaukus
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    |dw:1354600492668:dw|

    • one year ago
  21. UnkleRhaukus
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    now look at the graph

    • one year ago
  22. iheartfood
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    okay so (h,k) center=(-4,2) and radius=2 ???

    • one year ago
  23. iheartfood
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    is that right?? but how do i get the value of a from that?

    • one year ago
  24. UnkleRhaukus
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    \[\left(x+\frac a2\right)^2+\left(y+\frac b2\right)^2=a^2+b^2-c\] \[\left(x-h\right)^2+\left(y-k\right)^2=r^2\]

    • one year ago
  25. iheartfood
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    so a= -4/2 ???

    • one year ago
  26. iheartfood
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    so a=-2 ?? is that right?? my answer is -2??

    • one year ago
  27. jim_thompson5910
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    center=(-4,2) ---> h = -4, k = 2 radius=2 ---> r = 2 (x-h)^2 + (y-k)^2 = r^2 (x-(-4))^2 + (y-2)^2 = 2^2 (x+4)^2 + (y-2)^2 = 4 Now get this equation into x^2 + y^2 + ax + by + c = 0 form

    • one year ago
  28. iheartfood
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    okay, lemme give this a shot hahaha idk how close ill get tho!!! x^2 + y^2 + 4^2 - 2^2 - 4 = 0 ??? idk haha :P

    • one year ago
  29. iheartfood
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    is that right?? or even close?? :/

    • one year ago
  30. jim_thompson5910
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    no that's not correct

    • one year ago
  31. jim_thompson5910
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    (x-h)^2 + (y-k)^2 = r^2 (x-(-4))^2 + (y-2)^2 = 2^2 (x+4)^2 + (y-2)^2 = 4 x^2 + 8x + 16 + y^2 - 4y + 16 = 4

    • one year ago
  32. jim_thompson5910
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    keep going

    • one year ago
  33. iheartfood
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    aww darn :( okay so i simplify that right?

    • one year ago
  34. jim_thompson5910
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    yes

    • one year ago
  35. iheartfood
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    kk so i get this? x^2 + 8x + 16 + y^2 - 4y + 16 = 4 = x^2 + y^2 +8x -4y +12 = 0 ??

    • one year ago
  36. jim_thompson5910
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    not quite

    • one year ago
  37. jim_thompson5910
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    16+16-4 isn't 12

    • one year ago
  38. iheartfood
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    ohhh theres another 16!! i didn't see that! my bad!!! so i get this?? x^2 + y^2 +8x -4y + 28 = 0

    • one year ago
  39. jim_thompson5910
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    yep

    • one year ago
  40. iheartfood
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    kk so a=8 then?? my answer is 8? :O

    • one year ago
  41. jim_thompson5910
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    yes

    • one year ago
  42. iheartfood
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    awesome :) thanks so much for all of ur help @UnkleRhaukus and @jim_thompson5910 :)

    • one year ago
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