\[\boxed{\mathbf{1068.еуроки - ответы\ на\ пятёрку}}\]
\[\textbf{а)}\ (x + 2)^{2} + y^{2} = 1;\]
\[x^{2} + y^{2} = 4.\]
\[\left\{ \begin{matrix} (x + 2)^{2} + y^{2} = 1 \\ x^{2} + y^{2} = 4\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ ( - )\]
\[(x + 2)^{2} - x^{2} = - 3\]
\[x^{2} + 4x + 4 - x^{2} = - 3\]
\[4x = - 7\]
\[x = - \frac{7}{4}.\]
\[y^{2} = 4 - x^{2} = 4 - \left( \frac{7}{4} \right)^{2} =\]
\[= 4 - \frac{49}{16} = \frac{64 - 49}{16} = \frac{15}{16}\]
\[y = \pm \frac{\sqrt{15}}{4}.\]
\[Ответ:две\ точки\ пересечения.\]
\[\textbf{б)}\ (x + 3)^{2} + y^{2} = 1;\]
\[x^{2} + y^{2} = 4.\ \]
\[\left\{ \begin{matrix} (x + 3)^{2} + y^{2} = 1 \\ x^{2} + y^{2} = 4\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ ( - )\]
\[(x + 3)^{2} - x^{2} = - 3\]
\[x^{2} + 6x + 9 - x^{2} = - 3\]
\[6x = - 12\]
\[x = - 2.\]
\[y^{2} = 4 - x^{2} = 4 - 4 = 0;\]
\[y = 0.\]
\[Ответ:одна\ точка\ \]
\[пересечения\ (касания).\]