Equation of the circumference II: general equation

A circumference with center $$C = (a, b)$$ and radius $$r$$ can be rewritten in light of the reduced equation as:

$$$(x-a)^2+(y-b)^2=r^2$$$

Developing the squares of the above mentioned equation we obtain:

$$$x^2+y^2-2ax-2by+a^2+b^2-r^2=0$$$

and doing the change $$A= -2a, \ \ B=-2b, \ \ C=a^2+b^2-r^2$$ in:

$$$x^2+y^2-2ax-2by+a^2+b^2-r^2=0$$$

the new equation is obtained:

$$$x^2+y^2+Ax+By+C=0$$$

This way we have found another analytical expression that defines the points of a circumference. This is the general equation of the circumference.

Let's see how to determine the radius and the center of a circumference from the general equation.

We can do the following:

$$$A=-2a, \ \ B=-2b, \ \ C=a^2+b^2-r^2$$$

We isolate these expressions in terms of $$a$$, $$b$$ and $$r$$. We have:

$$$\displaystyle a=-\frac{A}{2}$$$ $$$b=-\frac{B}{2}$$$ $$$r^2=a^2+b^2-C=\Big(-\frac{A}{2}\Big)^2+\Big(-\frac{B}{2}\Big)^2-C=\frac{A^2+B^2-4C}{4}$$$

And since we know that, in the limited expression, $$(a, b)$$ is the center and $$r$$ the radius, given a general equation:

$$$x^2+y^2+Ax+Bx+C=0$$$

the center of such a circumference is the point $$\displaystyle \Big(-\frac{A}{2},-\frac{B}{2}\Big)$$ and the radius is $$\displaystyle r=\sqrt{\frac{A^2+B^2-4C}{4}}$$.

Let's now see the inverse process, that is to say:

Let's see what happens when the circumference is centred on the origin and we want to write its general equation:

Since $$(0, 0)$$ is the center we have: $$a=0$$ and $$b=0$$ for which reason,

$$$\left.{\begin{matrix} {0=a=-\frac{A}{2}} \\ {0=b=-\frac{B}{2}} \end{matrix}}\right \}\Longrightarrow{\left \{ {\begin{matrix} {A=0}\\{B=0}\end{matrix}}\right . }$$$

So that in the general equation, only quadratic terms and independent terms will exist, that is to say:

$$$x^2+y^2+C=0$$$

Moving the constant term to the other side we obtain:

$$$x^2+y^2=-C$$$

where we know that:

$$$C=a^2+b^2-r^2=-r^2$$$

since the supposed center was $$(0, 0)$$.

Note that for the circunference centered at the origin both equations are very similar.

Let's see an example:

Summing up we have:

Considering the circumference: $$(x-a)^2+(y-b)^2=r^2$$

Then the center is the point of the plane $$(a,b)$$ and the radius is $$r$$.

Considering the circumference: $$x^2+y^2+Ax+By+C=0$$

Then the center is at the point of the plane $$\displaystyle \Big(-\frac{A}{2},-\frac{B}{2}\Big)$$ and the radius is $$\displaystyle r=\sqrt{\Big(\frac{A}{2}\Big)^2+\Big(\frac{B}{2}\Big)^2-C}$$