Giải thích các bước giải:
a,
Gọi \(O\left( {a;b} \right)\) là tâm đường tròn ngoại tiếp tam giác ABC.
Suy ra \(OA = OB = OC = R\)
Ta có:
\(\begin{array}{l}
\overrightarrow {AO} = \left( {a + 1;b - 1} \right) \Rightarrow OA = \sqrt {{{\left( {a + 1} \right)}^2} + {{\left( {b - 1} \right)}^2}} \\
\overrightarrow {BO} \left( {a - 5;b - 2} \right) \Rightarrow OB = \sqrt {{{\left( {a - 5} \right)}^2} + {{\left( {b - 2} \right)}^2}} \\
\overrightarrow {CO} \left( {a - 3;b - 6} \right) \Rightarrow OC = \sqrt {{{\left( {a - 3} \right)}^2} + {{\left( {b - 6} \right)}^2}} \\
OA = OB = OC\\
\Rightarrow \left\{ \begin{array}{l}
\sqrt {{{\left( {a + 1} \right)}^2} + {{\left( {b - 1} \right)}^2}} = \sqrt {{{\left( {a - 5} \right)}^2} + {{\left( {b - 2} \right)}^2}} \\
\sqrt {{{\left( {a + 1} \right)}^2} + {{\left( {b - 1} \right)}^2}} = \sqrt {{{\left( {a - 3} \right)}^2} + {{\left( {b - 6} \right)}^2}}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{a^2} + 2a + 1 + {b^2} - 2b + 1 = {a^2} - 10a + 25 + {b^2} - 4b + 4\\
{a^2} + 2a + 1 + {b^2} - 2b + 1 = {a^2} - 6a + 9 + {b^2} - 12b + 36
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
12a + 2b = 27\\
8a + 10b = 43
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
a = \frac{{23}}{{13}}\\
b = \frac{{75}}{{26}}
\end{array} \right. \Rightarrow O\left( {\frac{{23}}{{13}};\,\,\frac{{75}}{{26}}} \right)
\end{array}\)
b,
Ta có:
\(\begin{array}{l}
\overrightarrow {AB} = \left( {6;1} \right) \Rightarrow AB = \sqrt {{6^2} + {1^2}} = \sqrt {37} \\
\overrightarrow {BC} = \left( { - 2;4} \right) \Rightarrow BC = \sqrt {{{\left( { - 2} \right)}^2} + {4^2}} = 2\sqrt 5 \\
\overrightarrow {CA} = \left( { - 4; - 5} \right) \Rightarrow CA = \sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 5} \right)}^2}} = \sqrt {41} \\
p = \frac{{AB + BC + CA}}{2}\\
{S_{ABC}} = \sqrt {p\left( {p - AB} \right)\left( {p - BC} \right)\left( {p - CA} \right)} = 13
\end{array}\)
