Question #118587
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).
1
Expert's answer
2020-06-01T18:28:08-0400

r=(xy1)r=\begin{pmatrix} x \\ y\\1 \end{pmatrix}

rT=(xy1)r^T=\begin{pmatrix} x & y&1 \end{pmatrix}

A=(10g01fgfc)A=\begin{pmatrix} 1 &0&g \\ 0 & 1&f\\ g&f&c \end{pmatrix}

rTAr=(xy1)(10g01fgfc)(xy1)==(x+gy+fxg+yf+c)(xy1)==(x2+xg+y2+yf+xg+yf+c)==(x2+2xg+y2+2yf+c)=0r^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

x2+2xg+g2g2+y2+2yf+f2f2+c=0(x+g)2+(y+f)2=g2+f2cx^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)(-g,-f) , radius R=g2+f2cR=\sqrt{g^2+f^2-c}



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