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Radius and coord of center of a circle |
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| Nov30-08, 08:01 PM | #1 |
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Radius and coord of center of a circle
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 |
| Nov30-08, 08:46 PM | #2 |
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Why not divide first by 9 to remove the pesky 9 in front of the square terms. Then try completing the squares.
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| Nov30-08, 08:48 PM | #3 |
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ok let me try, thanks
then its all fractions: x² - 2/3x + y² + 4/3 = 22/9 |
| Nov30-08, 08:54 PM | #4 |
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Radius and coord of center of a circle
Right. Now complete the square of x2 - 2/3x and y2 + 4/3.
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| Nov30-08, 09:02 PM | #5 |
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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 |
| Dec1-08, 06:16 AM | #6 |
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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.
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| Dec1-08, 03:07 PM | #7 |
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Recognitions:
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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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