Đáp án:

$\begin{array}{l}
Tâm\,I\left( {x;y} \right)\\
 \Rightarrow {d_{I - d}} = IA = IB\\
 \Rightarrow \frac{{\left| {x + 3y - 17} \right|}}{{\sqrt {1 + {3^2}} }} = \sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {y - 3} \right)}^2}} \\
 = \sqrt {{{\left( {x - 4} \right)}^2} + {{\left( {y - 1} \right)}^2}} \\
 \Rightarrow \left\{ \begin{array}{l}
\left| {x + 3y - 17} \right| = \sqrt {10} .\sqrt {{x^2} + 4x + 4 + {y^2} - 6y + 9} \\
{x^2} + 4x + 4 + {y^2} - 6y + 9 = {x^2} - 8x + 16\\
 + {y^2} - 2y + 1
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
{\left( {x + 3y - 17} \right)^2} = 10.\left( {{x^2} + {y^2} + 4x - 6y + 13} \right)\\
12x - 4y = 4
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
y = 3x - 1\\
{\left( {x + 9x - 3 - 17} \right)^2} = 10.{\left( {x + 2} \right)^2} + 10.{\left( {3x - 1 - 3} \right)^2}
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
y = 3x + 1\\
x = 1
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
y = 4\\
x = 1
\end{array} \right.\\
 \Rightarrow I\left( {1;4} \right)\\
 \Rightarrow R = IA = \sqrt {{{\left( {1 + 2} \right)}^2} + {{\left( {4 - 3} \right)}^2}}  = \sqrt {10} \\
 \Rightarrow \left( C \right):{\left( {x - 1} \right)^2} + {\left( {y - 4} \right)^2} = 10
\end{array}$