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:

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

Given, $3 x^{2}+3 y^{2}+6 x-4 y-1=0 \Rightarrow x^{2}+y^{2}+2 x-\frac{4}{3} y-\frac{1}{3}=0$

Comparing with (i) we see that the equation represents a circle with $2 \mathrm{~g}=2 \Rightarrow \mathrm{g}$ $=1,2 f=-\frac{4}{3} \Rightarrow f=-\frac{2}{3}$ and $c=-\frac{1}{3}$.

Centre $(-g,-f)=\left\{-1,-\left(-\frac{2}{3}\right)\right\}$

$=\left(-1, \frac{2}{3}\right)$

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

$=\sqrt{1^{2}+\left(-\frac{2}{3}\right)^{2}-\left(-\frac{1}{3}\right)}$

$=\sqrt{1+\frac{4}{9}+\frac{1}{3}}=\sqrt{\frac{16}{9}}=\frac{4}{3}$