Đáp án: $\left( C \right):{\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 1$ hoặc $\left( C \right):{\left( {x - 5} \right)^2} + {\left( {y + 5} \right)^2} = 25$
Giải thích các bước giải:
$\begin{array}{l}
I\left( {a;b} \right) \Rightarrow ptdt\left( C \right):{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {R^2}\\
*)d\left( {I,{\rm{Ox}}} \right) = d\left( {y;Oy} \right) \Leftrightarrow \left| a \right| = \left| b \right| = R \Leftrightarrow \left[ \begin{array}{l}
a = b\\
a = - b
\end{array} \right.\\
*)M\left( {2; - 1} \right) \in \left( C \right) \Rightarrow {\left( {2 - a} \right)^2} + {\left( { - 1 - b} \right)^2} = {R^2} = {a^2}\left( 1 \right)\\
+ TH1:a = b\\
\left( 1 \right) \Leftrightarrow {\left( {2 - a} \right)^2} + {\left( { - 1 - a} \right)^2} = {a^2} \Leftrightarrow {a^2} - 2a + 5 = 0(vn)\\
+ TH2:a = - b\\
\left( 1 \right) \Leftrightarrow {\left( {2 - a} \right)^2} + {\left( { - 1 + a} \right)^2} = {a^2} \Leftrightarrow {a^2} - 6a + 5 = 0 \Leftrightarrow \left[ \begin{array}{l}
a = 1 \Rightarrow b = - 1\\
a = 5 \Rightarrow b = - 5
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left( C \right):{\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 1\\
\left( C \right):{\left( {x - 5} \right)^2} + {\left( {y + 5} \right)^2} = 25
\end{array} \right.\\
\end{array}$
Vậy $\left( C \right):{\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 1$ hoặc $\left( C \right):{\left( {x - 5} \right)^2} + {\left( {y + 5} \right)^2} = 25$
