Question

Find the centre and radius of circle :

$2x^2+2y^2-x=0$

Answer 1

The equation of the given circle is $2x^2+2y^2-x=0$

$(2x^2-x)+2y^2=0$

$2\left[\right(x^2-\frac{x}{2})+y^2]=0$

$[x^2-2x(\frac{1}{4})+(\frac{1}{4}{)}^2]+y^2-(\frac{1}{4}{)}^2=0$

$(x-\frac{1}{4}{)}^2+(y-0{)}^2=(\frac{1}{4}{)}^2$

Thus, the centre of the given circle is $(\frac{1}{4},0)$ while its radius is $\frac{1}{4}$.