Show that the equation

The general equation of a conic is as follows

$a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0$ where $a, b, c, f, g, h$ are constants

For a circle, a = b and h = 0.

The equation becomes:

$x^{2}+y^{2}+2 g x+2 f y+c=0 \ldots$ (i)

Given, $x^{2}+y^{2}+x-y=0$

Comparing with (i) we see that the equation represents a circle with $2 \mathrm{~g}=1$

$\Rightarrow g=\frac{1}{2}, 2 f=-1 \Rightarrow f=-\frac{1}{2}$ and $c=0$

Centre $(-g,-f)=\left\{-\frac{1}{2},-\left(-\frac{1}{2}\right)\right\}$

$=\left(-\frac{1}{2}, \frac{1}{2}\right)$

Radius $=\sqrt{g^{2}+f^{2}-c}$

$=\sqrt{\frac{1^{2}}{2}+\left(-\frac{1^{2}}{2}\right)-0}$

$=\sqrt{\frac{1}{4}+\frac{1}{4}}=\sqrt{\frac{1}{2}}$