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If $\mathrm{z}=\mathrm{x}+\mathrm{iy}$ and the point $\mathrm{P}$ in the Argand plane represents $z$, then the locus of $z$ satisfying the equation $|z-2|+|z-2 i|$ $=4$ is
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$3 x^2+2 x y+3 y^2-8 x-8 y=0$
$\begin{aligned} & \text z=x+i y \\ & |z-2|+|z-2 i|=4 \\ & |(x-2)+i y|+|x+(y-2) i|=4 \\ & \Rightarrow \quad \sqrt{(x-2)^2+y^2}+\sqrt{x^2+(y-2)^2}=4 \\ & \Rightarrow \quad(x-2)^2+y^2=\left[4-\sqrt{x^2+(y-2)^2}\right]^2 \\ & \Rightarrow x^2+y^2-4 x+4 \\ & \quad=16+x^2+y^2+4-4 y-8 \sqrt{x^2+(y-2)^2} \\ & \Rightarrow 4 y-4 x-16=-8 \sqrt{x^2+(y-2)^2} \\ & \Rightarrow \quad(y-x-4)^2=4\left(x^2+(y-2)^2\right) \\ & \Rightarrow y^2+x^2+16-2 x y+8 x-8 y \\ & =4 x^2+4 y^2+16-16 y \\ & \Rightarrow 3 x^2+3 y^2+2 x y-8 x-8 y=0 .\end{aligned}$
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