Lập phương trình tiếp tuyến \(\Delta \) của đường tròn \(\left( C \right):\,\,{\left( {x + 2} \right)^2} + {\left( {y + 1} \right)^2} = 4,\) biết tiếp tuyến \(\Delta //d:\,\,\,x - 2y + 6 = 0.\)
A.\(\left[ \begin{array}{l}{\Delta _1}:\,\,2x - y + 2\sqrt 5 = 0\\{\Delta _2}:\,\,\,2x - y - 2\sqrt 5 = 0\end{array} \right.\)
B.\(\left[ \begin{array}{l}{\Delta _1}:\,\,\,x + 2y + 2\sqrt 5 = 0\\{\Delta _2}:\,\,\,x + 2y - 2\sqrt 5 = 0\end{array} \right.\)
C.\(\left[ \begin{array}{l}{\Delta _1}:\,\,\,x - 2y + 2\sqrt 5 = 0\\{\Delta _2}:\,\,\,x - 2y - 2\sqrt 5 = 0\end{array} \right.\)
D.\(\left[ \begin{array}{l}{\Delta _1}:\,\,\,2x + y + 2\sqrt 5 = 0\\{\Delta _2}:\,\,\,2x + y - 2\sqrt 5 = 0\end{array} \right.\)
A.\(\left[ \begin{array}{l}{\Delta _1}:\,\,2x - y + 2\sqrt 5 = 0\\{\Delta _2}:\,\,\,2x - y - 2\sqrt 5 = 0\end{array} \right.\)
B.\(\left[ \begin{array}{l}{\Delta _1}:\,\,\,x + 2y + 2\sqrt 5 = 0\\{\Delta _2}:\,\,\,x + 2y - 2\sqrt 5 = 0\end{array} \right.\)
C.\(\left[ \begin{array}{l}{\Delta _1}:\,\,\,x - 2y + 2\sqrt 5 = 0\\{\Delta _2}:\,\,\,x - 2y - 2\sqrt 5 = 0\end{array} \right.\)
D.\(\left[ \begin{array}{l}{\Delta _1}:\,\,\,2x + y + 2\sqrt 5 = 0\\{\Delta _2}:\,\,\,2x + y - 2\sqrt 5 = 0\end{array} \right.\)
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Hóa Học lớp 12
