Lời giải chi tiết:

Nối \(AK \cap \left( C \right) = M;\,\,BK \cap \left( C \right) = N\)

\( \Rightarrow M,N\) là trung điểm của cung BC và AC.

Giả sử \(M\left( {a;b} \right)\,\,\left\{ \begin{array}{l}\overrightarrow {AM}  = \left( {a + 1;b - 2} \right)\\\overrightarrow {AK}  = \left( {3; - 1} \right)\end{array} \right.\)

\(\overrightarrow {AM} //\overrightarrow {AK}  \Rightarrow \frac{{a + 1}}{3} = \frac{{b - 2}}{{ - 1}}\,\,\left( 1 \right)\)

\(I{A^2} = I{M^2} \Leftrightarrow {\left( {a - \frac{3}{2}} \right)^2} + {\left( {b - 2} \right)^2} = {\left( {\frac{3}{2} + 1} \right)^2} + {\left( {2 - 2} \right)^2}\,\,\,\left( 2 \right)\)

Giải hệ \(\left\{ \begin{array}{l}\left( 1 \right)\\\left( 2 \right)\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}M\left( {\frac{7}{2};\frac{1}{2}} \right)\\M\left( { - 1;2} \right)\,\,\left( {ktm} \right)\end{array} \right.\)

\(\widehat{{{B}_{1}}}=\frac{sd\overset\frown{MN}}{2}=\frac{sd\overset\frown{NC}+sd\overset\frown{CM}}{2}=\frac{sd\overset\frown{BM}+sd\overset\frown{AN}}{2}=\widehat{{{K}_{1}}}\)

\( \Rightarrow \Delta BMK\) cân tại M.

Giả sử \(B\left( {x;y} \right) \Rightarrow \left\{ \begin{array}{l}IA = IB\\MB = MK\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\left( {x - \frac{3}{2}} \right)^2} + {\left( {y - 2} \right)^2} = \frac{{25}}{4}\\{\left( {x - \frac{7}{2}} \right)^2} + {\left( {y - \frac{1}{2}} \right)^2} = \frac{5}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}B\left( {4;2} \right)\\B\left( {\frac{5}{8}; - \frac{5}{2}} \right)\,\,\left( {ktm} \right)\end{array} \right.\)