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Câu hỏi: Cho số phức z thỏa mãn $\left| z+3 \right|=5$ và $\left| z-2i \right|=\left| z-2-2i \right|$. Khi đó $\left| z \right|$ bằng
A. $\left| z \right|=10.$
B. $\left| z \right|=17.$
C. $\left| z \right|=\sqrt{17}.$
D. $\left| z \right|=\sqrt{10}.$
A. $\left| z \right|=10.$
B. $\left| z \right|=17.$
C. $\left| z \right|=\sqrt{17}.$
D. $\left| z \right|=\sqrt{10}.$
Đặt $z=x+iy;x,y\in R.$
Theo bài ra ta có:$\left\{ \begin{aligned}
& {{\left( x+3 \right)}^{2}}+{{y}^{2}}=25 \\
& {{x}^{2}}+{{\left( y-2 \right)}^{2}}={{\left( x-2 \right)}^{2}}+{{\left( y-2 \right)}^{2}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\left( x+3 \right)}^{2}}+{{y}^{2}}=25 \\
& -4x+4=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{y}^{2}}=9 \\
& x=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& y=\pm 3 \\
& x=1 \\
\end{aligned} \right.. $Vậy$ \left| z \right|=\sqrt{10}.$
Theo bài ra ta có:$\left\{ \begin{aligned}
& {{\left( x+3 \right)}^{2}}+{{y}^{2}}=25 \\
& {{x}^{2}}+{{\left( y-2 \right)}^{2}}={{\left( x-2 \right)}^{2}}+{{\left( y-2 \right)}^{2}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\left( x+3 \right)}^{2}}+{{y}^{2}}=25 \\
& -4x+4=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{y}^{2}}=9 \\
& x=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& y=\pm 3 \\
& x=1 \\
\end{aligned} \right.. $Vậy$ \left| z \right|=\sqrt{10}.$
Đáp án D.