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Radius and coord of center of a circle

  1. Nov 30, 2008 #1
    1. The problem statement, all variables and given/known data
    9x² + 9y² - 6x + 12y - 22 = 0


    2. Relevant equations



    3. The attempt at a solution

    i got this far but cant finish i dont know what to do:

    3(3x² - 2x + 1) + 3(3y² + 4y + 4) = 40
     
  2. jcsd
  3. Nov 30, 2008 #2
    Why not divide first by 9 to remove the pesky 9 in front of the square terms. Then try completing the squares.
     
  4. Nov 30, 2008 #3
    ok let me try, thanks

    then its all fractions:

    x² - 2/3x + y² + 4/3 = 22/9
     
  5. Nov 30, 2008 #4
    Right. Now complete the square of x2 - 2/3x and y2 + 4/3.
     
  6. Nov 30, 2008 #5
    i don't know what half of 2/3 is or 4/3 cuz if i know that then i can square them and thats it

    i calculated :

    (x² - 2/3x + 1/9) + (y² + 4/3 + 2/3) = 22/9

    is this it
     
    Last edited: Nov 30, 2008
  7. Dec 1, 2008 #6
    That should be 4/3y by the way and the 2/3 term should be squared. You're almost done. Now write (x2 - 2/3x + 1/9) as (x - ?)2 and do the same with the other term.
     
  8. Dec 1, 2008 #7

    Mentallic

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    Homework Helper

    It's quite simple really. half of 2/3 => [tex]\frac{2}{3}\div \frac{2}{1}=\frac{2}{3}x\frac{1}{2}[/tex]
    now multiply both numerator and denominators together => [tex]\frac{2}{6}=\frac{1}{3}[/tex]

    Similarly for 4/3. So half of 2/3 is 1/3 and half of 4/3 is 2/3. Sounds logical right?
     
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