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 suppose that they give us the circumference: $$x^2+y^2-2x+4y-4=0$$ then we see that it is centred at the point:
$$\displaystyle \Big(-\frac{A}{2},-\frac{B}{2}\Big)=\Big(-\frac{-2}{2},-\frac{4}{2}\Big)=(1,-2)$$
and has radius:
$$\displaystyle r=\sqrt{\frac{A^2+B^2-4C}{4}}=\sqrt{\frac{(-2)^2+4^2-4\cdot(- 4)}{4}}=$$
$$=\displaystyle\sqrt{\frac{4+16+16}{4}}=\sqrt{\frac{36}{4}}=\frac{6}{2}=3$$
Let's now see the inverse process, that is to say:
Giving the general equation of the circumference that has, for example, radius $4$ and center $(-5, 6)$.
We write the reduced equation:
$$(x-a)^2+(y-b)^2=r^2 \Rightarrow (x+5)^2+(y-6)^2=4^2 $$
developing the squares we have:
$$(x+5)^2+(y-6)^2=4^2 \Rightarrow x^2+10x+25+y^2-12y+36=16$$
If we rearrange it and add all the independent terms, we obtain the general equation of the above mentioned circumference, which is:
$$x^2+10x+25+y^2-12y+36=16$$
$$x^2+y^2+10x-12y+25+36=16$$
$$x^2+y^2+10x-12y+45=0$$
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:
Circumference centred on the origin and radius $7$.
Reduced equation: $x^2+y^2=7^2$
General equation: $x^2+y^2+C=0$ where $C=-7^2 \Longrightarrow x^2+y^2-7^2=0$
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 $r$.
$(x-8)^2+(y+3)^2=1$ has center at $(8,-3)$ and radius $1$.
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 $\displaystyle r=\sqrt{\Big(\frac{A}{2}\Big)^2+\Big(\frac{B}{2}\Big)^2-C}$
$x^2+y^2+x-5y-2=0$ has center $\displaystyle \Big(\frac{-1}{2},\frac{5}{2}\Big)$ and radius
$$\displaystyle r=\sqrt{\Big(\frac{1}{2}\Big)^2+\Big(\frac{-5}{2}\Big)^2-(-2)}=\sqrt{\frac{1+25+8}{4}}=\sqrt{\frac{34}{4}}=\sqrt{\frac{17}{2}}$$