## 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
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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.
 ok let me try, thanks then its all fractions: x² - 2/3x + y² + 4/3 = 22/9

## Radius and coord of center of a circle

Right. Now complete the square of x2 - 2/3x and y2 + 4/3.
 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
 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.

Recognitions:
Homework Help
 Quote by lucifer_x 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
It's quite simple really. half of 2/3 => $$\frac{2}{3}\div \frac{2}{1}=\frac{2}{3}x\frac{1}{2}$$
now multiply both numerator and denominators together => $$\frac{2}{6}=\frac{1}{3}$$

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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