Answer
$\left(x-\dfrac{1}{2}\right)^2+(y+2)^2=\dfrac{9}{2}$
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:
$$\begin{align*}
2x^2+2y^2-2x+8y&=\dfrac{1}{2}\\
x^2+y^2-x+4y&=\dfrac{1}{4}\\
\left(x^2-x+\dfrac{1}{4}\right)+(y^2+4y+4)-\dfrac{1}{4}-4&=\dfrac{1}{4}\\
\left(x-\dfrac{1}{2}\right)^2+(y+2)^2&=\dfrac{9}{2}.
\end{align*}$$
The equation
$$\left(x-\dfrac{1}{2}\right)^2+(y+2)^2=\dfrac{9}{2}\tag2$$
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 $r$:
$$\begin{align*}
h&=\dfrac{1}{2}\\
k&=-2\\
r&=\sqrt{\dfrac{9}{2}}=\dfrac{3}{\sqrt 2}=\dfrac{3\sqrt 2}{2}.
\end{align*}$$
Use the center and the radius to graph the circle:
