Trong không gian với hệ trục Oxyz, cho mặt cầu

C

- Hướng dẫn giải

\(\begin{array}{l}\left( S \right):{\left( {x - 1} \right)^2} + {\left( {y - 2} \right)^2} + {\left( {z + 2} \right)^2} = 25\\I\left( {1;2; - 2} \right);R = 5\end{array}\)

Dễ thấy \(A\left( {1; - 3;0} \right);B\left( {3;1;4} \right) \in d\) nên:

\(\begin{array}{l}\left( P \right):a\left( {x - 1} \right) + b\left( {y + 3} \right) + cz = 0\\a.\left( {3 - 1} \right) + b\left( {1 + 3} \right) + c.4 = 0\\ \Rightarrow a = - 2b - 2c\\ \Rightarrow \left( P \right):\left( { - 2b - 2c} \right)\left( {x - 1} \right) + b\left( {y + 3} \right) + cz = 0\end{array}\)

(P) tiếp xúc với (S) khi:

 \(\begin{array}{l}{d_{I/\left( P \right)}} = R \Leftrightarrow \frac{{\left| {\left( { - 2b - 2c} \right)\left( {1 - 1} \right) + b\left( {2 + 3} \right) + c\left( { - 2} \right)} \right|}}{{\sqrt {{{\left( {2b - 2c} \right)}^2} + {b^2} + {c^2}} }} = 5\\ \Leftrightarrow \frac{{\left| {5b - 2c} \right|}}{{\sqrt {5{b^2} + 8bc + 5{c^2}} }} = 5\\ \Leftrightarrow 25{b^2} - 20bc + 4{c^2} = 25\left( {5{b^2} + 8bc + 5{c^2}} \right)\\ \Leftrightarrow 100{b^2} + 220bc + 121{c^2} = 0\\ \Leftrightarrow {\left( {10b + 11c} \right)^2} = 0 \Leftrightarrow b = \frac{{ - 11}}{{10}}c\\ \Rightarrow \left( P \right):\left( { - 2.\left( {\frac{{ - 11}}{{10}}} \right) - 2} \right)\left( {x - 1} \right) - \frac{{11}}{{10}}\left( {y + 3} \right) + z = 0\\ \Rightarrow \left( P \right):2x - 11y + 10z - 35 = 0\end{array}\)