Given: the equation of the circle, we need to rewrite it in the form $(x-h)^2+(y-k)^2=r^2$, so that we can determine the centre as (h,k) and radius as $r$
Given $2x^2+2y^2-x = 0$
$\quad \quad 2(x^2-\large\frac{1}{2}$$x) + 2y^2 = 0 \rightarrow (x^2-\large\frac{1}{2}$$x) + 2y^2 = 0$
$\quad \quad (x^2-2x\large\frac{1}{4}$$ + \large\frac{1}{4}^2) + y^2 - \large\frac{1}{4}^2 = 0$
$\quad \quad (x-\large\frac{1}{4} ) ^2 $$ + (y-0)^2 = \large\frac{1}{4}^2$
$\Rightarrow$ centre = $(\large\frac{1}{4}$$,\;0)$ and radius = $\large\frac{1}{4}$