Equation of Circle/Cartesian/Corollary 1

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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

\(\displaystyle A \paren {x^2 + y^2} + B x + C y + D\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle x^2 + y^2 + \frac B A x + \frac C A y\) \(=\) \(\displaystyle - \frac D A\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle x^2 + 2 \frac B {2 A} x + \frac {B^2} {4 A^2} + y^2 + 2 \frac C {2 A} y + \frac {C^2} {4 A^2}\) \(=\) \(\displaystyle \frac {B^2} {4 A^2} + \frac {C^2} {4 A^2} - \frac D A\)
\(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \paren {x + \frac B {2 A} }^2 + \paren {y + \frac C {2 A} }^2\) \(=\) \(\displaystyle \frac {B^2} {4 A^2} + \frac {C^2} {4 A^2} - \frac D A\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {B^2} {4 A^2} + \frac {C^2} {4 A^2} - \frac {4 A D} {4 A^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {4 A^2} \paren {B^2 + C^2 - 4 A D}\)

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

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.

$\blacksquare$