Question 214690
The last equation should be {{{(x-5)^2+(y+3)^2=9}}}, since {{{y^2+6y+9}}} factors to {{{(y+3)^2}}}. We can take this equation and then rewrite {{{9}}} as {{{3^2}}} and {{{y+3}}} to {{{y-(-3)}}} to get {{{(x-5)^2+(y-(-3))^2=3^2}}}



Recall that the general equation of a circle is {{{(x-h)^2+(y-k)^2=r^2}}} where (h,k) is the center and 'r' is the radius.



Looking at {{{(x-5)^2+(y-(-3))^2=3^2}}} (which is in the circle form described above), we see that {{{h=5}}}, {{{k=-3}}} and {{{r=3}}} (I'm just matching the two forms).



So the center is (5,-3) and the radius is 3 units.



So you just had some minor errors (such as the sign differences and forgetting to take the square root of 9)