[UPLIFTJEE]

Let z be a complex number such that $\left|\frac{z-2 i}{z+i}\right|=2, z \neq-i$. Then z lies on the circle of radius 2 and centre
[JEE (Main) 25 Jan 2023 S II], [JEE (Main) 25 Jan 2023 S II]
(A) $(0,0)$
(B) (0, -2)
(C) $(2,0)$
(D) $(0,2)$

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[UPLIFTJEE]

Solution

$(\mathrm{z}-2 \mathrm{i})(\overline{\mathrm{z}}+2 \mathrm{i})=4(\mathrm{z}+\mathrm{i})(\overline{\mathrm{z}}-\mathrm{i})$

\[ \begin{aligned} & z \bar{z}+4+2 i(z-\bar{z})=4(z \bar{z}+1+i(\bar{z}-z)) \\ & 3 z \bar{z}-6 i(z-\bar{z})=0 \Rightarrow x^{2}+y^{2}-2 i(2 i y)=0 \Rightarrow x^{2}+y^{2}+4 y=0 \end{aligned} \]

[UPLIFTJEE]

Answer

B

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