Question:
Find the centre and radius of the circle $2 x^{2}+2 y^{2}-x=0$
Solution:
The equation of the given circle is $2 x^{2}+2 y^{2}-x=0$.
$2 x^{2}+2 y^{2}-x=0$
$\Rightarrow\left(2 x^{2}-x\right)+2 y^{2}=0$
$\Rightarrow 2\left[\left(x^{2}-\frac{x}{2}\right)+y^{2}\right]=0$
$\Rightarrow\left\{x^{2}-2 x\left(\frac{1}{4}\right)+\left(\frac{1}{4}\right)^{2}\right\}+y^{2}-\left(\frac{1}{4}\right)^{2}=0$
$\Rightarrow\left(x-\frac{1}{4}\right)^{2}+(y-0)^{2}=\left(\frac{1}{4}\right)^{2}$ which is of the form $(x-h)^{2}+(y-k)^{2}=r^{2}$, where $h=\frac{1}{4}, k=0$, and $r=\frac{1}{4}$
Thus, the centre of the given circle is $\left(\frac{1}{4}, 0\right)$, while its radius is $\frac{1}{4}$.