Answer
$a)$ Yes, the equation represents a circle
$b)$ Center: $\Big(\dfrac{1}{2},-2\Big);$ radius: $\dfrac{3\sqrt{2}}{2}$
Work Step by Step
$2x^{2}+2y^{2}-2x+8y=\dfrac{1}{2}$
$a)$
Take out common factor $2$ from the left side:
$2(x^{2}+y^{2}-x+4y)=\dfrac{1}{2}$
Take the $2$ to divide the right side:
$x^{2}+y^{2}-x+4y=\dfrac{1}{(2)(2)}$
$x^{2}+y^{2}-x+4y=\dfrac{1}{4}$
On the left side, group the terms with $x$ together and the terms with $y$ together:
$(x^{2}-x)+(y^{2}+4y)=\dfrac{1}{4}$
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, $b=-1$ and for the second group, $b=4$
$\Big[x^{2}-x+\Big(\dfrac{-1}{2}\Big)^{2}\Big]+\Big[y^{2}+4y+\Big(\dfrac{4}{2}\Big)^{2}\Big]=\dfrac{1}{4}+\Big(\dfrac{-1}{2}\Big)^{2}+\Big(\dfrac{4}{2}\Big)^{2}$
$\Big(x^{2}-x+\dfrac{1}{4}\Big)+\Big(y^{2}+4y+4\Big)=\dfrac{1}{4}+\dfrac{1}{4}+4$
$\Big(x^{2}-x+\dfrac{1}{4}\Big)+\Big(y^{2}+4y+4\Big)=\dfrac{9}{2}$
Now, each group is a perfect square trinomial. Factor them both:
$\Big(x-\dfrac{1}{2}\Big)^{2}+\Big(y+2\Big)^{2}=\dfrac{9}{2}$
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:
$\Big(x-\dfrac{1}{2}\Big)^{2}+\Big(y+2\Big)^{2}=\dfrac{9}{2}$
radius: $r=\sqrt{\dfrac{9}{2}}=\dfrac{3}{\sqrt{2}}=\dfrac{3\sqrt{2}}{2}$
Center: $\Big(\dfrac{1}{2},-2\Big);$ radius: $\dfrac{3\sqrt{2}}{2}$