SOLUTION: I haven't taken algebra in 4 years I just need help refreshing. The problem says to find the center and the radius of the graph of the circle. 9x^2+9y^2-6y-17=0

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Question 24796: I haven't taken algebra in 4 years I just need help refreshing. The problem says to find the center and the radius of the graph of the circle.
9x^2+9y^2-6y-17=0
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
First, you need to put your equation 9x%5E2+%2B+9y%5E2+-+6y+-+17+=+0 into the standard form for a circle with centre at (h, k) and radius r. %28x-h%29%5E2+%2B+%28y-k%29%5E2+=+r%5E2
To accomplish this, you will need to "complete the square" in the y-terms. The x-term is already squared so it does not need to be changed. Here are the steps:
1) 9x%5E2+%2B+9y%5E2+-+6y+-+17+=+0 Add 17 to both sides of the equation.
2) 9x%5E2+%2B+9y%5E2+-+6y+=+17 Divide through by 9.
3) %28x%5E2%29+%2B+%28y%5E2+-+%282%2F3%29y%29+=+17%2F9 Complete the square in the y-terms by adding the square of half the y-coefficient (that's %28%281%2F2%29%282%2F3%29%29%5E2+=+1%2F9) to both sides of the equation.
4) %28x%5E2%29+%2B+%28y%5E2+-+%282%2F3%29y+%2B+1%2F9%29+=+17+%2B+1%2F9 Simplify and factor the y-group.
5) %28x%5E2%29+%2B+%28y-1%2F3%29%5E2+=+154%2F9 Rewrite the x-term as %28x+-+0%29%5E2
6) %28x+-+0%29%5E2+%2B+%28y+-+1%2F3%29%5E2+=+154%2F9 Compare this with the standard form:
7) %28x+-+h%29%5E2+%2B+%28y+-+k%29%5E2+=+r%5E2
You can see that the centre: (h, k) is (0, 1/3) and the radius, r, is sqrt%28154%2F9%29+=+%281%2F3%29sqrt%28154%29