\[\boxed{\mathbf{1001.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[A(3;0);\]
\[B( - 1;2);\]
\[A\ и\ B \in окружности;\]
\[O \in y = x + 2.\]
\[\mathbf{Найти:}\]
\[уравнение\ окружности.\]
\[\mathbf{Решение.}\]
\[1)\ R = AO =\]
\[= \sqrt{(3 - x)^{2} + (0 - y)^{2}} =\]
\[= \sqrt{(3 - x)^{2} + y^{2}};\]
\[R = BO =\]
\[= \sqrt{( - 1 - x)^{2} + (2 - y)^{2}}.\]
\[2)\ (3 - x)^{2} + y^{2} =\]
\[= ( - 1 - x)^{2} + (2 - y)^{2}\]
\[9 - 6x + x^{2} + y^{2} =\]
\[= 1 + 2x + x^{2} + 4 - 4y + y^{2}\]
\[9 - 6x = 2x - 4y + 5\]
\[\left. \ 8x - 4y - 4 = 0\ \ \ \ \ \ \right|\ :4\]
\[2x - y - 1 = 0.\]
\[3)\ \left\{ \begin{matrix} 2x - y - 1 = 0 \\ y = x + 2\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} 2x - x - 2 - 1 = 0 \\ y = x + 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x = 3 \\ y = 5 \\ \end{matrix} \Longrightarrow O(3;5). \right.\ \]
\[4)\ R = AO = \sqrt{(3 - 3)^{2} + 5^{2}} =\]
\[= \sqrt{25} = 5.\]
\[5)\ Уравнение\ окружности:\]
\[(x - 3)^{2} + (y - 5)^{2} = 25.\]
\[Ответ:(x - 3)^{2} + (y - 5)^{2} = 25.\]