Trong mặt phẳng tọa độ Oxy, cho tam giác ABC có A(-4;1),B(2;4)

* Hướng dẫn giải

Gọi I( x; y). Ta có $\left\{\begin{array}{l}\overrightarrow{AI}=\left(x+4;y-1\right)\\ \overrightarrow{BI}=\left(x-2;y-4\right)\\ \overrightarrow{CI}=\left(x-2;y+2\right)\end{array}\right..$

Do I là tâm đường tròn ngoại tiếp tam giác ABC nên $IA=IB=IC\iff \left\{\begin{array}{l}I{A}^2=I{B}^2\\ I{B}^2=I{C}^2\end{array}\right.$

$\begin{array}{l}\iff \left\{\begin{array}{l}{\left(x+4\right)}^2+{\left(y-1\right)}^2={\left(x-2\right)}^2+{\left(y-4\right)}^2\\ {\left(x-2\right)}^2+{\left(y-4\right)}^2={\left(x-2\right)}^2+{\left(y+2\right)}^2\end{array}\right.\iff \left\{\begin{array}{l}{\left(x+4\right)}^2+{\left(y-1\right)}^2={\left(x-2\right)}^2+{\left(y-4\right)}^2\\ {\left(y-4\right)}^2={\left(y+2\right)}^2\end{array}\right.\\ \iff \left\{\begin{array}{l}{\left(x+4\right)}^2={\left(x-2\right)}^2+{(1-4)}^2\\ y=1\end{array}\iff \left\{\begin{array}{c}{x}^2+\text{}8x+16=x^2-4x+4+\text{}9\\ y=1\end{array}\right.\right.\iff \left\{\begin{array}{l}x=-\frac{1}{4}\\ y=1\end{array}\right.\end{array}$.

Chọn B.