\(\left( C \right)\) có tâm I(1;2) bán kính R=3
a) \(I' = {Q_{\left( {O;{{90}^0}} \right)}}\left( I \right) \Rightarrow \left\{ \begin{array}{l}{x_{I'}} = - {y_I} = - 2\\{y_{I'}} = {x_I} = 1\end{array} \right. \Rightarrow I'\left( { - 2;1} \right)\)
Đường tròn \(\left( {C'} \right)\) có tâm \(I'\left( { - 2;1} \right)\) bán kính R’=R=3 nên có phương trình:
\({\left( {x + 2} \right)^2} + {\left( {y - 1} \right)^2} = 9\)
b)
\(\begin{array}{l}M'' = {Q_{\left( {O; - {{90}^0}} \right)}}\left( M \right) \Rightarrow \left\{ \begin{array}{l}{x_{M''}} = {y_M} = 1\\{y_{M''}} = - {x_M} = - 1\end{array} \right. \Rightarrow M''\left( {1; - 1} \right)\\M' = {T_{\overrightarrow v }}\left( {M''} \right) \Leftrightarrow \left\{ \begin{array}{l}{x_{M'}} = {x_{M''}} + 3 = 1 + 3 = 4\\{y_{M'}} = {y_{M''}} + 4 = - 1 + 4 = 3\end{array} \right.\\ \Rightarrow M'\left( {4;3} \right)\end{array}\)
c)
\(\begin{array}{l}N'' = {V_{\left( {O;2} \right)}}\left( N \right) \Rightarrow \left\{ \begin{array}{l}{x_{N''}} = 2{x_N} = 2.2 = 4\\{y_{N''}} = 2{y_N} = 2.3 = 6\end{array} \right. \Rightarrow N''\left( {4;6} \right)\\N' = {T_{\overrightarrow v }}\left( {N''} \right) \Leftrightarrow \left\{ \begin{array}{l}{x_{N'}} = {x_{N''}} + 3 = 4 + 3 = 7\\{y_{N'}} = {y_{N''}} + 4 = 6 + 4 = 10\end{array} \right.\\ \Rightarrow N'\left( {7;10} \right)\end{array}\)
d) Gọi d’:2x-3y+c=0
Lấy \(A\left( {0; - 2} \right) \in d\),
\(\begin{array}{l}A' = {V_{\left( {O; - 2} \right)}}\left( A \right) \Rightarrow \left\{ \begin{array}{l}{x_{A'}} = - 2{x_A} = - 2.0 = 0\\{y_{A'}} = - 2{y_A} = - 2.\left( { - 2} \right) = 4\end{array} \right.\\ \Rightarrow A'\left( {0;4} \right) \in d' \Leftrightarrow 2.0 - 3.4 + c = 0 \Leftrightarrow c = 12\\ \Rightarrow d':2x - 3y + 12 = 0\end{array}\)
