$A(2;-1);C(-5;4)$
I là trung điểm AC
$\Rightarrow I(\dfrac{x_A+x_C}{2};\dfrac{y_A+y_C}{2})\Rightarrow I(\dfrac{-3}{2};\dfrac{3}{2})$
$\vec{IA}=(x_A-x_I;y_A-y_I)=(\dfrac{7}{2};\dfrac{-5}{2})$
$IA=\sqrt{(x_A-x_I)^2+(y_A-y_I)^2}=\sqrt{\dfrac{37}{2}}$Đường kính AC suy ra tâm $I(\dfrac{-3}{2};\dfrac{3}{2})$ bán kính $R=IA=\sqrt{\dfrac{37}{2}}\\ \Rightarrow (C): (x+\dfrac{3}{2})^2+(y-\dfrac{3}{2})^2=\dfrac{37}{2}$
$\vec{AC}=(x_C-x_A;y_C-y_A)=(-7;5)$
