Question 38903
Hello!
We need to rewrite your equation to get the form:

{{{(x-h)^2 + (y-k)^2 = r^2}}}

Once we have the equation in this for, we know that the center is at (h, k)

You have:
{{{x^2+y^2-6y=7}}}

{{{(x-0)^2 + y^2 -6y -7 = 0}}}

{{{(x-0)^2 + y^2 -6y + 9 - 9 -7 = 0}}}

{{{(x-0)^2 + y^2 -6y + 9 - 16 = 0}}}

{{{(x-0)^2 + (y-3)^2 - 16 = 0}}}

{{{(x-0)^2 + (y-3)^2 = 16}}}

{{{(x-0)^2 + (y-3)^2 = 4^2}}}

So the center of this circle is at (h, k) = (0, 3)


I hope this helps!
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