b: Đặt I(x;y)
I là tâm đường tròn ngoại tiếp ΔABC nên IA=IB=IC
=>\(\left\{{}\begin{matrix}IA^2=IB^2\\IB^2=IC^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(1-x\right)^2+\left(-2-y\right)^2=\left(3-x\right)^2+\left(2-y\right)^2\\\left(3-x\right)^2+\left(2-y\right)^2=\left(7-x\right)^2+\left(4-y\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-2x+1+y^2+4y+4=x^2-6x+9+y^2-4y+4\\x^2-6x+9+y^2-4y+4=x^2-14x+49+y^2-8y+16\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2x+4y+5=-6x-4y+13\\-6x-4y+13=-14x-8y+65\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+8y=8\\8x+4y=52\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}8x+16y=16\\8x+4y=52\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}16y-4y=16-52\\x+2y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12y=-36\\x=2-2y\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-3\\x=2-2y=2-2\cdot\left(-3\right)=2+6=8\end{matrix}\right.\)
Vậy: I(8;-3)
