Question 693439
This is solved by completing the squares:

{{{x^2+y^2-x+2y+1=0}}}

{{{x^2-x +(y^2+2y+1)=0}}}

{{{x^2-2*(1/2)*x+(1/2)^2-(1/2)^2+(y+1)^2=0}}}

{{{(x-1/2)^2+(y+1)^2=(1/2)^2}}}

Compare to {{{(x-x[0])^2+(y-y[0])^2=r^2}}} where {{{x[0]}}} ,{{{y[0]}}} are the coordinates of the circle's center and {{{r}}} is the radius.

The center of the circle is at {{{x[0]=1/2}}}, {{{y[0]=-1}}} and the radius is {{{r=1/2}}}.