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$\mathrm{z} \overline{\mathrm{z}}+(3-\mathrm{i}) \mathrm{z}+(3+\mathrm{i}) \overline{\mathrm{z}}+1=0$ represents a circle with
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centre $(-3,-1)$ and radius 3
Let $z=x+$ iy
$\bar{z}=x-i y$
$\Rightarrow \quad \mathrm{zz}+(3-\mathrm{i}) \mathrm{z}+(3+\mathrm{i}) \overline{\mathrm{z}}+1=0$
$\Rightarrow \quad(\mathrm{x}+\mathrm{iy})(\mathrm{x}-\mathrm{iy})+(3-\mathrm{i})(\mathrm{x}+\mathrm{iy})+(3+\mathrm{i})(\mathrm{x}-\mathrm{iy})+1=0$
$\Rightarrow \quad x^{2}+y^{2}+6 x+2 y+1=0$
$\Rightarrow \quad(x+3)^{2}-9+(y+1)^{2}-1+1=0$
$\Rightarrow \quad(x+3)^{2}+(y+1)^{2}=(3)^{2}$
Centre $(-3,-1)$ radius $=3$
$\bar{z}=x-i y$
$\Rightarrow \quad \mathrm{zz}+(3-\mathrm{i}) \mathrm{z}+(3+\mathrm{i}) \overline{\mathrm{z}}+1=0$
$\Rightarrow \quad(\mathrm{x}+\mathrm{iy})(\mathrm{x}-\mathrm{iy})+(3-\mathrm{i})(\mathrm{x}+\mathrm{iy})+(3+\mathrm{i})(\mathrm{x}-\mathrm{iy})+1=0$
$\Rightarrow \quad x^{2}+y^{2}+6 x+2 y+1=0$
$\Rightarrow \quad(x+3)^{2}-9+(y+1)^{2}-1+1=0$
$\Rightarrow \quad(x+3)^{2}+(y+1)^{2}=(3)^{2}$
Centre $(-3,-1)$ radius $=3$
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