The given equation of circle is, $2x^2 + 2y^2$ – x = 0 ∴ $x^2+y^2-\frac{x}{2}$ = 0 ⇒ $\left(x^2-\frac{x}{2}\right)+y^2$ = 0 ⇒ $\left(x^2-\frac{x}{2}+\left(\frac{1}{4}\right)^2\right)+y^2$ = $0+\left(\frac{1}{4}\right)^2$ ⇒ $\left(x-\frac{1}{4}\right)^2+y^2$ = $\left(\frac{1}{4}\right)^2$ Comparing it with $(x - h)^2 + (y - k)^2$ = $r^2$, we have h = $\frac{1}{4}$ , k = 0 and r = $\frac{1}{4}$ Thus co-ordinates of the centre are $\left(\frac{1}{4},0\right)$ and radius is $\frac{1}{4}$.