Trong mặt phẳng tọa độ Oxy, cho ta...

B

- Hướng dẫn giải

Gọi I(a; b) là tâm đường tròn ngoại tiếp tam giác ABC.

$\left\{\begin{array}{c}A{I}^2=B{I}^2\\ A{I}^2=C{I}^2\end{array}\right.\iff \left\{\begin{array}{c}{\left(a-0\right)}^2+{\left(b-2\right)}^2={\left(a+2\right)}^2+{\left(b-8\right)}^2\\ {\left(a-0\right)}^2+{\left(b-2\right)}^2={\left(a+3\right)}^2+{\left(b-1\right)}^2\end{array}\right.$

$\iff \left\{\begin{array}{c}{a}^2+b^2-4b+4=a^2+4a+4+b^2-16b+64\\ a^2+b^2-4b+4=a^2+6a+9+b^2-2b+1\end{array}\right.$

$\left\{\begin{array}{c}4a-12b=-64\\ 6a+2b=-6\end{array}\iff \right.\left\{\begin{array}{c}a-3b=-16\\ 3a+b=-3\end{array}\right.$

$\iff \left\{\begin{array}{c}a=-\frac{5}{2}\\ b=\frac{9}{2}\end{array}\right.$

Chọn B.