Equation of Circle/Cartesian/Formulation 2

Theorem
The equation:
 * $A \left({x^2 + y^2}\right) + B x + C y + D = 0$

is the equation of a circle with radius $R$ and center $\left({a, b}\right)$, where:
 * $R = \dfrac 1 {2 A} \sqrt {B^2 + C^2 - 4 A D}$
 * $\left({a, b}\right) = \left({\dfrac B {2 A}, \dfrac C {2 A} }\right)$

provided $B^2 + C^2 \ge 4 A D$.

Proof
This last expression is non-negative $B^2 + C^2 \ge 4 A D$.

In such a case $\dfrac 1 {4 A^2} \left({B^2 + C^2 - 4 A D}\right)$ is in the form $R^2$ and so:


 * $\left({x + \dfrac B {2 A} }\right)^2 + \left({y + \dfrac C {2 A} }\right)^2 = \dfrac 1 {4 A^2} \left({B^2 + C^2 - 4 A D}\right)$

is in the form:
 * $\left({x - a}\right)^2 + \left({y - b}\right)^2 = R^2$

Hence the result from Equation of Circle: Cartesian.