Answer to Question #118587 in Analytic Geometry for Nii Laryea

Question

If r = x
y
1
1 0 g
and A = 0 1 f
g f c
show that r^T Ar = 0 is the equation of the
circle in R^2 with centre (−g, −f) and radius square root(g^2 + f^2 − c).

Answer 1

$r=\begin{pmatrix} x \\ y\\1 \end{pmatrix}$

$r^T=\begin{pmatrix} x & y&1 \end{pmatrix}$

$A=\begin{pmatrix} 1 &0&g \\ 0 & 1&f\\ g&f&c \end{pmatrix}$

$r^TAr=\begin{pmatrix} x & y&1 \end{pmatrix}\begin{pmatrix} 1 &0&g \\ 0 & 1&f\\ g&f&c \end{pmatrix}\begin{pmatrix} x \\ y\\1 \end{pmatrix} =\\ =\begin{pmatrix} x+g&y+f&xg+yf+c \end{pmatrix}\begin{pmatrix} x \\ y\\1 \end{pmatrix}=\\ =(x^2+xg+y^2+yf+xg+yf+c)=\\ =(x^2+2xg+y^2+2yf+c)=0$

$x^2+2xg+g^2-g^2+y^2+2yf+f^2-f^2+c=0\\ (x+g)^2+(y+f)^2=g^2+f^2-c$

The circle: centre $(-g,-f)$ , radius $R=\sqrt{g^2+f^2-c}$