Equation of Circle/Cartesian/Formulation 2

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

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

provided:
 * $A > 0$
 * $B^2 + C^2 \ge 4 A D$

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

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


 * $\paren {x + \dfrac B {2 A} }^2 + \paren {y + \dfrac C {2 A} }^2 = \dfrac 1 {4 A^2} \paren {B^2 + C^2 - 4 A D}$

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

Hence the result from Equation of Circle: Cartesian.