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Let $z=x+i y$ Where $x, y$ are real variables $i=\sqrt{-1}$. If
$|2 z-1|=|z-2|$, then the point $z$ describes:
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$|2 z-1|=|z-2|$, then the point $z$ describes:
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Verified Answer
The correct answer is:
A circle
$|2 z-1|=|z-2|$
$|2(\mathrm{x}+\mathrm{iy})-1|=|\mathrm{x}+\mathrm{iy}-2|$
$|(2 x-1)+2 y i|=|(x-2)+i y|$
$\sqrt{(2 \mathrm{x}-1)^{2}+\mathrm{y}^{2}}=\sqrt{(\mathrm{x}-2)^{2}+\mathrm{y}^{2}}$
Squaring both sides $4 x^{2}+1-4 x+4 y^{2}=x^{2}+4-4 x+y^{2}$
$\Rightarrow 3 x^{2}+3 y^{2}=3$
$\Rightarrow x^{2}+y^{2}=1$
It is the equation of a circle. $\therefore$ The point $z$ describes a circle.
$|2(\mathrm{x}+\mathrm{iy})-1|=|\mathrm{x}+\mathrm{iy}-2|$
$|(2 x-1)+2 y i|=|(x-2)+i y|$
$\sqrt{(2 \mathrm{x}-1)^{2}+\mathrm{y}^{2}}=\sqrt{(\mathrm{x}-2)^{2}+\mathrm{y}^{2}}$
Squaring both sides $4 x^{2}+1-4 x+4 y^{2}=x^{2}+4-4 x+y^{2}$
$\Rightarrow 3 x^{2}+3 y^{2}=3$
$\Rightarrow x^{2}+y^{2}=1$
It is the equation of a circle. $\therefore$ The point $z$ describes a circle.
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