Cho tam giác ABC có A(3; -3), B(-3...

B

- Hướng dẫn giải

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

Ta có: AI = BI = CI $\iff$ AI² =  BI² = CI²

$\left\{\begin{array}{c}A{I}^2=B{I}^2\\ B{I}^2=C{I}^2\end{array}\right.\iff \left\{\begin{array}{c}{\left(a-3\right)}^2+{\left(b+3\right)}^2={\left(a+3\right)}^2+{\left(b-5\right)}^2\\ {\left(a+3\right)}^2+{\left(b-5\right)}^2={\left(a-3\right)}^2+{\left(b-5\right)}^2\end{array}\right.$

$\begin{array}{l}\iff \left\{\begin{array}{c}{a}^2-6a+\text{}9+\text{}{b}^2+\text{}6b+\text{}9=a^2+\text{}6a+9+\text{}{b}^2-10b+25\\ a^2+6a+\text{}9+\text{}{b}^2-10b+\text{}25=a^2-6a+9+b^2-10b+\text{}25\end{array}\right.\\ \iff \left\{\begin{array}{c}-12a+16b=16\\ 12a=0\end{array}\right.\iff \left\{\begin{array}{c}a=0\\ b=1\end{array}\right.\end{array}$

Vậy tâm I(0; 1).

Chọn B.