\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2+\left(y-3\right)^2=1\\\left(x-1\right)\left(y-3\right)-\left(x-1\right)-\left(y-3\right)=1\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\left(x-1\right)+\left(y-3\right)=a\\\left(x-1\right)\left(y-3\right)=b\end{matrix}\right.\) với \(a^2\ge4b\) ta được hệ:
\(\left\{{}\begin{matrix}a^2-2b=1\\b-a=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2-2b=1\\b=a+1\end{matrix}\right.\)
\(\Rightarrow a^2-2\left(a+1\right)=1\Leftrightarrow a^2-2a-3=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-1\Rightarrow b=0\\a=3\Rightarrow b=4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left(x-1\right)\left(y-3\right)=0\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=2\\y=3\Rightarrow x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2+\left(y-3\right)^2=1\\2\left(x-1\right)\left(y-3\right)-2\left(x+y\right)=-6\end{matrix}\right.\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y-3\right)^2+2\left(x-1\right)\left(y-3\right)+1-2\left(x+y\right)=-4\)
\(\Leftrightarrow\left(x+y-4\right)^2-2\left(x+y-4\right)=4\)
\(\Leftrightarrow\left(x+y-5\right)^2=5\Leftrightarrow\left[{}\begin{matrix}x=5+\sqrt{5}-y\\x=5-\sqrt{5}-y\end{matrix}\right.\)
Xét từng TH thay vào tính được x,y
Có j sai bn tính lại nha
