Question

Find the cente and radius for the following circle.
$2x^2+2y^2-x=0$

Answer 1

The given equation of circle is
$2x^2+2y^2-x=0\therefore x^2+y^2-\frac{x}{2}=0\Rightarrow (x^2-\frac{x}{2})+y^2=0\Rightarrow [x^2-\frac{x}{2}+{\left(\frac{1}{4}\right)}^2]+y^2=0+{\left(\frac{1}{4}\right)}^2\Rightarrow {(x-\frac{1}{4})}^2+y^2={\left(\frac{1}{4}\right)}^2$
comparing it with $(x-h{)}^2+(y-k{)}^2=r^2,$ we have
$h=\frac{1}{4},k=0andr=\frac{1}{4}$
Thus co-ordinates of the centre is $(\frac{1}{4},0)$

Radius is $\frac{1}{4}$