Question 24796
First, you need to put your equation {{{9x^2 + 9y^2 - 6y - 17 = 0}}} into the standard form for a circle with centre at (h, k) and radius r. {{{(x-h)^2 + (y-k)^2 = r^2}}}
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^2 + 9y^2 - 6y - 17 = 0}}} Add 17 to both sides of the equation.
2) {{{9x^2 + 9y^2 - 6y = 17}}} Divide through by 9.
3) {{{(x^2) + (y^2 - (2/3)y) = 17/9}}} Complete the square in the y-terms by adding the square of half the y-coefficient (that's {{{((1/2)(2/3))^2 = 1/9}}}) to both sides of the equation.
4) {{{(x^2) + (y^2 - (2/3)y + 1/9) = 17 + 1/9}}} Simplify and factor the y-group.
5) {{{(x^2) + (y-1/3)^2 = 154/9}}} Rewrite the x-term as {{{(x - 0)^2}}}
6) {{{(x - 0)^2 + (y - 1/3)^2 = 154/9}}} Compare this with the standard form:
7) {{{(x - h)^2 + (y - k)^2 = r^2}}}

You can see that the centre: (h, k) is (0, 1/3) and the radius, r, is {{{sqrt(154/9) = (1/3)sqrt(154)}}}