Trong không gian với hệ tọa độ Oxyz...

D

- Hướng dẫn giải

Đáp án D.

Gọi O là trung điểm của AB, suy ra $O\left(0;0;0\right)$  .

Ta có $\overrightarrow{AB}=\left(-2x_0;0;0\right),\overrightarrow{OC}=\left(0;1;0\right)$$\Rightarrow \overrightarrow{AB}.\overrightarrow{OC}=0\Rightarrow AB\perp OC$.

Gắn hệ trục tọa độ Oxyz như hình vẽ bên. Với $A\left(x_0;0;0\right),B\left(-x_0;0;0\right),C\left(0;1;0\right)$,$B\text{'}\left(-x_0;0;4-x_0\right)$,$A\text{'}\left(x_0;0;4-x_0\right)$, $C\text{'}\left(0;1;4-x_0\right)$ do  $x_0+y_0=4$ và $0<x_0,y_0<4$.

 Có $\overrightarrow{AC\text{'}}=\left(-x_0;1;4-x_0\right),\overrightarrow{B\text{'}C}=\left(x_0;1;x_0-4\right)$$\Rightarrow \left[\overrightarrow{AC\text{'}},\overrightarrow{B\text{'}C}\right]=\left(2x_0-8;0;-2x_0\right)$

$$\begin{gathered} \overrightarrow{AC}=\left(-x_0;1;0\right) \\ \Rightarrow \left[\overrightarrow{AC\text{'}},\overrightarrow{B\text{'}C}\right].\overrightarrow{AC} \\ =-x_0\left(2x_0-8\right)=-2x_0\left(x_0-4\right) \end{gathered}$$

$\begin{array}{l}\Rightarrow d\left(AC\text{'};B\text{'}C\right)\\ =\frac{\left|\left[\overrightarrow{AC\text{'}},\overrightarrow{B\text{'}C}\right].\overrightarrow{AC}\right|}{\left|\left[\overrightarrow{AC\text{'}},\overrightarrow{B\text{'}C}\right]\right|}\\ =\frac{\left|2x_0\left(x_0-4\right)\right|}{\sqrt{4{\left(4-x_0\right)}^2+4x_0^2}}\\ =\frac{x_0\left(4-x_0\right)}{\sqrt{{\left(4-x_0\right)}^2+x_0^2}}\end{array}$

 do $x_0\in \left(0;4\right)$.

Với $0<x_0<4$  , ta có ${\left(4-x_0\right)}^2+x_0^2\stackrel{AM-GM}{\ge }$$2\sqrt{{\left(4-x_0\right)}^2{x}_0^2}=2x_0\left(4-x_0\right)$.

Như vậy $d\left(AC\text{'};B\text{'}C\right)=\frac{x_0\left(4-x_0\right)}{\sqrt{{\left(4-x_0\right)}^2+x_0^2}}$$\le \frac{x_0\left(4-x_0\right)}{2x_0\left(4-x_0\right)}=\frac{1}{2}$.

Dấu “=” xảy ra khi $x_0=4-x_0\iff x_0=2=y_0$  .

Khi đó $A\left(2;0;0\right),B\left(-2;0;0\right),$$C\left(0;1;0\right),B\text{'}\left(-2;0;2\right)$. Giả sử phương trình mặt cầu ngoại tiếp lăng trụ $ABC.A\text{'}B\text{'}C\text{'}$ là .

Ta có hệ phương trình sau:

 $\begin{array}{l}\left\{\begin{array}{l}{2}^2+0^2+0^2-2a.2-2b.0-2c.0+d=0\\ {\left(-2\right)}^2+0^2+0^2-2a\left(-2\right)-2b.0-2c.0+d=0\\ 0^2+1^2+0^2-2a.0-2b.1-2c.0+d=0\\ {\left(-2\right)}^2+0^2+2^2-2a\left(-2\right)-2b.0-2c.2+d=0\end{array}\right.\\ \iff \left\{\begin{array}{l}4a-d=4\\ 4a+d=-4\\ 2b-d=1\\ 4a-4c+d=-8\end{array}\right.\\ \iff \left\{\begin{array}{l}a=0\\ b=-\frac{3}{2}\\ c=1\\ d=-4\end{array}\right.\end{array}$

Vậy mặt cầu (S) có tâm $I\left(0;-\frac{3}{2};1\right)$ và bán kính 

$$\begin{gathered} R=\sqrt{a^2+b^2+c^2-d} \\ =\frac{\sqrt{29}}{2} \end{gathered}$$