I(a;b)
[tex]PTDT: (x-a)^2+(y-b)^2=R^2[/tex]
(C) đi qua A(-1,2) , B(-2,3)
[tex]\Rightarrow \left\{\begin{matrix} & (a+1)^2+(b-2)^2=R^2 & \\ & (a+2)^2+(b-3)^2=R^2 & \end{matrix}\right.\Rightarrow (a+1)^2+(b-2)^2=(a+2)^2+(b-3)^2\\\Rightarrow 2a-2b+8=0[/tex]
I thuộc 3x-y+10=0 nên 3a-b+10=0
[tex]\left\{\begin{matrix} & 2a-2b+8=0 & \\ & 3a-b+10=0 & \end{matrix}\right.\Rightarrow a=-3,b=1\\\Rightarrow I(-3;1)\Rightarrow IA=\sqrt{5}\\\Rightarrow (x+3)^2+(y-1)^2=5[/tex]