Đáp án:
\({\left( {x + 1} \right)^2} + {\left( {y - 1} \right)^2} = 1\)
Giải thích các bước giải:
\(\begin{array}{l}
\left( C \right):\,\,{\left( {x - 2} \right)^2} + {\left( {y - 2} \right)^2} = 4\,\,co\,\,tam\,\,I\left( {2;2} \right),\,\,ban\,\,kinh\,\,R = 2\\
{V_{\left( {O;\frac{1}{2}} \right)}}\left( I \right) = I'\left( {x';y'} \right) \Rightarrow \left\{ \begin{array}{l}
x' = \frac{1}{2}.2 = 1\\
y' = \frac{1}{2}.2 = 1
\end{array} \right. \Rightarrow I'\left( {1;1} \right)\,\,va\,\,R' = \frac{1}{2}.2 = 1\\
\Rightarrow {V_{\left( {O;\frac{1}{2}} \right)}}\left( C \right) = \left( {C'} \right)\,\,co\,\,tam\,\,I'\left( {1;1} \right),\,\,ban\,\,kinh\,\,R' = 1.\\
{Q_{\left( {O;{{90}^0}} \right)}}\left( {I'} \right) = I''\left( { - 1;1} \right),\,\,R'' = R' = 1\\
\Rightarrow {Q_{\left( {O;{{90}^0}} \right)}}\left( {C'} \right) = \left( {C''} \right)\,\,co\,\,tam\,\,I''\left( { - 1;1} \right),\,\,ban\,\,kinh\,\,R'' = 1\\
\Rightarrow \left( {C''} \right):\,\,{\left( {x + 1} \right)^2} + {\left( {y - 1} \right)^2} = 1.
\end{array}\)
