Gọi \(M\left(x;y;0\right)\) \(\Rightarrow OM^2=x^2+y^2\)
\(d^2\left(M;\left(\alpha\right)\right)=\frac{\left(x+2y+4\right)^2}{9}\) ; \(d^2\left(M;\left(\beta\right)\right)=\frac{\left(2x-2y-13\right)^2}{9}\)
\(\left(x+2y+4\right)^2=\left(2x-2y-13\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2y+4=2x-2y-13\\x+2y+4=-2x+2y+13\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4y+17\\3x=9\Rightarrow x=3\end{matrix}\right.\)
Th1: \(\left\{{}\begin{matrix}x=3\\x^2+y^2=\frac{\left(x+2y+4\right)^2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\9y^2+81=4y^2+28y+49\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\5y^2-28y+32=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}M\left(3;4;0\right)\\M\left(3;\frac{8}{5};0\right)\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x=4y+17\\x^2+y^2=\frac{\left(x+2y+4\right)^2}{9}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4y+17\\\left(4y+17\right)^2+y^2=\left(2y+7\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4y+17\\13y^2+108y+240=0\end{matrix}\right.\) (vô nghiệm)
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