Question

Find the centre and radius of the circle $2x^2+2y^2-x=0$

Answer 1

Given $2x^2+2y^2-x=0$
$\Rightarrow (2x^2-x)+2y^2=0$
$\Rightarrow 2[(x^2-\frac{x}{2})+y^2]=0$
$\Rightarrow \{x^2-2x\left(\frac{1}{4}\right)+{\left(\frac{1}{4}\right)}^2\}+y^2-{\left(\frac{1}{4}\right)}^2=0$
$\Rightarrow {(x-\frac{1}{4})}^2+{(y-0)}^2={\left(\frac{1}{4}\right)}^2$
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 $(\frac{1}{4},0)$ while its radius is $\frac{1}{4}$.