جواب کار در کلاس صفحه 44 درس 2 هندسه دوازدهم (آشنایی با مقاطع مخروطی)

الف

\(\begin{array}{*{20}{l}}{{x^2} + {y^2} - 4x - 6y = 3}\\{}\\{ \Rightarrow \left\{ \begin{array}{l}O(2\;,\;3)\\\\r = \frac{{\sqrt {16 + 36 + 12} }}{2} = 4\end{array} \right.}\\{}\\{{x^2} + {y^2} - 10x - 14y + 73 = 0}\\{}\\{ \Rightarrow \left\{ \begin{array}{l}O'(5\;,\;7)\\\\r' = \frac{{\sqrt {100 + 196 - 292} }}{2} = 1\end{array} \right.}\\{}\\\begin{array}{l} \Rightarrow d = OO' = \\\\\sqrt {{{(5 - 2)}^2} + {{(7 - 3)}^2}} = 4\;\end{array}\\{}\\\begin{array}{l}r + r' = 4 + 1 = 5\\\\\left| {r - r'} \right| = 4 - 1 = 3\end{array}\\{}\\{ \Rightarrow \left| {r - r'} \right| < d < r + r'}\end{array}\)

دو دایره متقاطع اند.

ب

\(\begin{array}{*{20}{l}}{{x^2} + {y^2} - 2x = 1}\\{}\\{ \Rightarrow O(1\;,\;0)\;,\;r = \frac{{\sqrt {4 + 0 + 4} }}{2} = \sqrt 2 }\\{}\\{{x^2} + {y^2} = 1 \Rightarrow O'(0\;,\;0)\;,\;r' = 1}\\{}\\\begin{array}{l} \Rightarrow d = OO' = \\\\\sqrt {{{(0 - 1)}^2} + {{(0 - 0)}^2}} = 1\end{array}\\{}\\{r + r' = 1 + \sqrt 2 = 5}\\{}\\{\left| {r - r'} \right| = \sqrt 2 - 1}\\{}\\{ \Rightarrow \left| {r - r'} \right| < d < r + r'}\end{array}\)

دو دایره متقاطع اند.

ج

\(\begin{array}{*{20}{l}}\begin{array}{l}{x^2} + {y^2} = 9\\\\ \Rightarrow \left\{ \begin{array}{l}O(0\;,\;0)\\\\r = 3\end{array} \right.\end{array}\\{}\\{{x^2} + {y^2} - 2x + 2y + 1 = 0}\\{}\\{ \Rightarrow \left\{ \begin{array}{l}O'(1\;,\; - 1)\\\\r' = \frac{{\sqrt {4 + 4 - 4} }}{2} = 1\end{array} \right.}\\{}\\\begin{array}{l} \Rightarrow d = OO' = \\\\\sqrt {{{(0 - 1)}^2} + {{(0 + 1)}^2}} = \sqrt 2 \end{array}\\{}\\\begin{array}{l}r + r' = 3 + 1 = 4\\\\\left| {r - r'} \right| = 3 - 1 = 2\end{array}\\{}\\{ \Rightarrow d < \left| {r - r'} \right|}\end{array}\)

دو دایره متداخل اند.

د

\(\begin{array}{*{20}{l}}\begin{array}{l}{x^2} + {y^2} = 4\\\\ \Rightarrow \left\{ \begin{array}{l}O(0\;,\;0)\\\\r = 2\end{array} \right.\end{array}\\{}\\{{x^2} + {y^2} - 8x - 4y + 19 = 0}\\{}\\{ \Rightarrow \left\{ \begin{array}{l}O'(4\;,\; - 3)\\\\r' = \frac{{\sqrt {64 + 16 - 76} }}{2} = 1\end{array} \right.}\\{}\\\begin{array}{l} \Rightarrow d = OO' = \\\\\sqrt {{{(4 - 0)}^2} + {{( - 2 - 0)}^2}} = \\\end{array}\\{\sqrt {20} \; = 2\sqrt 5 }\\{}\\\begin{array}{l}r + r' = 2 + 1 = 3\\\\\left| {r - r'} \right| = 2 - 1 = 1\end{array}\\{}\\{ \Rightarrow r + r' < d}\end{array}\)

دو دایره متخارج اند.