Answer
$(x+1)^2+(y-3)^2=1$
Work Step by Step
The equation of a circle is:
$$(x-h)^2+(y-k)^2=r^2\tag1$$
a) We complete the squares in the given equation:
$$\begin{align*}
x^2+y^2+2x-6y+9&=0\\
(x^2+2x+1)+(y^2-6y+9)-1&=0\\
(x+1)^2+(y-3)^2&=1.
\end{align*}$$
The equation
$$(x+1)^2+(y-3)^2=1\tag2$$
is in the form presented in Eq. $(1)$, therefore it represents a circle.
b) We determine the coordinates $(h,k)$ of the center and the radius $r$:
$$\begin{align*}
h&=-1\\
k&=3\\
r&=1.
\end{align*}$$
Use the center and the radius to graph the circle:
