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Câu hỏi: Cho đường thẳng d $\dfrac{x-1}{2}=\dfrac{y-2}{1}=\dfrac{z-3}{2}$ và hai mặt phẳng $\left( {{P}_{1}} \right):x+2y+2z-2=0;\left( {{P}_{2}} \right):2x+y+2z-1=0$. Mặt cầu có tâm I nằm trên d và tiếp xúc với 2 mặt phẳng $\left( {{P}_{1}} \right),\left( {{P}_{2}} \right)$, có phương trình.
A. $\left( S \right):{{\left( x+1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}+{{\left( z+3 \right)}^{2}}=9$.
B. $\left( S \right):{{\left( x+1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}+{{\left( z+3 \right)}^{2}}=9$.
C. $\left( S \right):{{\left( x-1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}+{{\left( z-3 \right)}^{2}}=3$.
D. $\left( S \right):{{\left( x-1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}+{{\left( z-3 \right)}^{2}}=9$.
A. $\left( S \right):{{\left( x+1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}+{{\left( z+3 \right)}^{2}}=9$.
B. $\left( S \right):{{\left( x+1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}+{{\left( z+3 \right)}^{2}}=9$.
C. $\left( S \right):{{\left( x-1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}+{{\left( z-3 \right)}^{2}}=3$.
D. $\left( S \right):{{\left( x-1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}+{{\left( z-3 \right)}^{2}}=9$.
• $I\in d\Rightarrow I\left( 2t+1;t+2;2t+3 \right)$
• Mặt cầu tiếp xúc với 2 mặt phẳng $\Leftrightarrow d\left( I;\left( {{P}_{1}} \right) \right)=d\left( {{I}_{2}};\left( {{P}_{2}} \right) \right)$
$\Leftrightarrow \left| 8t+9 \right|=\left| 9t+9 \right|\Leftrightarrow \left[ \begin{aligned}
& 8t+9=9t+9 \\
& 8t-9=-9t-9 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& t=0 \\
& t=\dfrac{-18}{17} \\
\end{aligned} \right.$
• $t=0\Rightarrow I\left( 1;2;3 \right);R=3\Rightarrow \left( S \right):{{\left( x-1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}+{{\left( z-3 \right)}^{2}}=9$.
• $t=-\dfrac{18}{17}\Rightarrow I\left( -\dfrac{19}{17};\dfrac{16}{17};\dfrac{15}{17} \right);R=\dfrac{3}{17}\Rightarrow \left( S \right):{{\left( x+\dfrac{19}{17} \right)}^{2}}+{{\left( y-\dfrac{16}{17} \right)}^{2}}+{{\left( z-\dfrac{15}{17} \right)}^{2}}=\dfrac{9}{289}$.
• Mặt cầu tiếp xúc với 2 mặt phẳng $\Leftrightarrow d\left( I;\left( {{P}_{1}} \right) \right)=d\left( {{I}_{2}};\left( {{P}_{2}} \right) \right)$
$\Leftrightarrow \left| 8t+9 \right|=\left| 9t+9 \right|\Leftrightarrow \left[ \begin{aligned}
& 8t+9=9t+9 \\
& 8t-9=-9t-9 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& t=0 \\
& t=\dfrac{-18}{17} \\
\end{aligned} \right.$
• $t=0\Rightarrow I\left( 1;2;3 \right);R=3\Rightarrow \left( S \right):{{\left( x-1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}+{{\left( z-3 \right)}^{2}}=9$.
• $t=-\dfrac{18}{17}\Rightarrow I\left( -\dfrac{19}{17};\dfrac{16}{17};\dfrac{15}{17} \right);R=\dfrac{3}{17}\Rightarrow \left( S \right):{{\left( x+\dfrac{19}{17} \right)}^{2}}+{{\left( y-\dfrac{16}{17} \right)}^{2}}+{{\left( z-\dfrac{15}{17} \right)}^{2}}=\dfrac{9}{289}$.
Đáp án D.