Answer
Center: $\Big(\dfrac{1}{2},-1\Big)$ $;$ Radius: $\dfrac{1}{2}$
Work Step by Step
$x^{2}+y^{2}-x+2y+1=0$
Take $1$ to the right side:
$x^{2}+y^{2}-x+2y=-1$
Group the terms with $x$ together and the terms with $y$ together:
$(x^{2}-x)+(y^{2}+2y)=-1$
Complete the square for each group by adding $\Big(\dfrac{b}{2}\Big)^{2}$ to each expression inside parentheses and also on the right side of the equation:
For the first group, $b=-1$, for the second group, $b=2$
$\Big[x^{2}-x+\Big(-\dfrac{1}{2}\Big)^{2}\Big]+\Big[y^{2}+2y+\Big(\dfrac{2}{2}\Big)^{2}\Big]=-1+\Big(-\dfrac{1}{2}\Big)^{2}+\Big(\dfrac{2}{2}\Big)^{2}$
$\Big(x^{2}-x+\dfrac{1}{4}\Big)+(y^{2}+2y+1)=-1+1+\dfrac{1}{4}$
$\Big(x^{2}-x+\dfrac{1}{4}\Big)+(y^{2}+2y+1)=\dfrac{1}{4}$
Factor the expressions inside parentheses, which are perfect square trinomials:
$\Big(x-\dfrac{1}{2}\Big)^{2}+(y+1)^{2}=\dfrac{1}{4}$
This is now the standard form of the equation of a circle, which is $(x-h)^{2}+(y-k)^{2}=r^{2}$
It can be seen from the obtained equation that $(h,k)=\Big(\dfrac{1}{2},-1\Big)$, so the center of the circle is the point $\Big(\dfrac{1}{2},-1\Big)$
The radius is:
$r^{2}=\dfrac{1}{4}$
$\sqrt{r^{2}}=\sqrt{\dfrac{1}{4}}$
$r=\dfrac{1}{2}$
The graph is:
