Gọi đường tròn tâm \(I\left(a;b\right)\Rightarrow d\left(I;d_1\right)=d\left(I;d_2\right)\)
\(\Rightarrow\dfrac{\left|3a-4b+1\right|}{5}=\dfrac{\left|4a+3b-7\right|}{5}\)
\(\Rightarrow\left[{}\begin{matrix}3a-4b+1=4a+3b-7\\3a-4b+1=-4a-3b+7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}a=-7b+8\\b=7a-6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}I\left(-7b+8;b\right)\\I\left(a;7a-6\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}IA^2=\left(-7b+6\right)^2+\left(b-3\right)^2\\IA^2=\left(a-2\right)^2+\left(7a-9\right)^2\end{matrix}\right.\)
\(IA^2=d^2\left(I;d_1\right)\Rightarrow\left[{}\begin{matrix}\left(-7b+6\right)^2+\left(b-3\right)^2=\left(b-1\right)^2\\\left(a-2\right)^2+\left(7a-9\right)^2=\left(a-1\right)^2\end{matrix}\right.\)
Giờ giải pt bậc 2 là được
