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, $x^{2}+y^{2}-4 x+6 y-5=0$

Comparing with (i) we see that the equation represents a circle with $2 \mathrm{~g}=-4 \Rightarrow \mathrm{g}=-2$, $2 f=6 \Rightarrow f=3$ and $c=-5$.

Centre $(-g,-f)=\{-(-2),-3\}$

$=(2,-3)$

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

$=\sqrt{(-2)^{2}+3^{2}-(-5)}$

$=\sqrt{4+9+5}=\sqrt{18}=3 \sqrt{2}$