Answer
$(x-1)^2+(y+3)^2=1$
Center: $(1,-3)$
Radius: $r=1$
Work Step by Step
$x^2+y^2-2x+6y+9=0$
Notice that $-2x=-2(1)x$ and that $1^2=1$
Notice that $6y=2(3)y$ and that $3^2=9$
$x^2-2x+1-1+y^2+6y+9-9+9=0$
$(x^2-2x+1)+(y^2+6y+9)=9-9+1$
$(x-1)^2+(y+3)^2=1$
The equation of a circle in standard form:
$(x-h)^2+(y-k)^2=r^2$ in which $(h,k)$ is the center and $r$ is the radius
$(x-1)^2+[y-(-3)]^2=1^2$
Center: $(1,-3)$
Radius: $r=1$
