Question

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

Answer 1

Given:$2x^2+2y^2=x$

$\Rightarrow x^2+y^2=\frac{x}{2}$

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

$\Rightarrow (x^2-2\times x\times \frac{1}{4}+\frac{1}{16})+{(y-0)}^2=\frac{1}{16}$

$\Rightarrow {(x-\frac{1}{4})}^2+{(y-0)}^2={\left(\frac{1}{4}\right)}^2$,which is comparable with ${(x-h)}^2+{(y-k)}^2=r^2$

Here $h=\frac{1}{4},k=0$ and $r=\frac{1}{4}$

Hence, the given circle has centre $(\frac{1}{4},0)$ and its radius$=\frac{1}{4}$