Answer
$a)$ Yes, the equation represents a circle
$b)$ Center: $(-1,3);$ radius: $1$
Work Step by Step
$x^{2}+y^{2}+2x-6y+9=0$
$a)$
Take $9$ to the right side:
$x^{2}+y^{2}+2x-6y=-9$
Group the terms with $x$ together and the terms with $y$ together:
$(x^{2}+2x)+(y^{2}-6y)=-9$
Complete the square for each group. To do so, add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation and do it for both groups. For the first group $b=2$ and for the second group, $b=-6$
$\Big[x^{2}+2x+\Big(\dfrac{2}{2}\Big)^{2}\Big]+\Big[y^{2}-6y+\Big(\dfrac{-6}{2}\Big)^{2}\Big]=-9+\Big(\dfrac{2}{2}\Big)^{2}+\Big(\dfrac{-6}{2}\Big)^{2}$
$(x^{2}+2x+1)+(y^{2}-6y+9)=-9+1+9$
$(x^{2}+2x+1)+(y^{2}-6y+9)=1$
Now, each group is a perfect square trinomial. Factor them both:
$(x+1)^{2}+(y-3)^{2}=1$
Since it was possible to represent the original equation in the standard form of the equation of a circle, then it represents a circle.
$b)$
The equation of a circle is $(x-h)^{2}+(y-k)^{2}=r^{2}$, where $(h,k)$ is the center of the circle and $r$ is its radius.
Obtain the center and the radius from the equation found in part $a$ of this problem:
$(x+1)^{2}+(y-3)^{2}=1$
Center: $(-1,3);$ radius: $1$
