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If $z=x+i y$, then the centre of the circle $\left|\frac{z-3}{z-2 i}\right|=2$, is
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Verified Answer
The correct answer is:
$\left(-1, \frac{8}{3}\right)$
We have,
$$
z=x+i y
$$
Given, $\left|\frac{z-3}{z-2 i}\right|=2$
$$
\begin{aligned}
& |(x-3)+i y|=2|x+(y-2) i| \\
& (x-3)^2+y^2=4\left(x^2+(y-2)^2\right. \\
& =x^2-6 x+9+y^2=4 x^2+4 y^2-16 y+16 \\
\Rightarrow & 3 x^2+3 y^2+6 x-16 y+7=0 \\
\Rightarrow & x^2+y^2+2 x-\frac{16}{3} y+\frac{7}{3}=0 \\
\therefore & \text { Centre of circle }\left(-1, \frac{8}{3}\right)
\end{aligned}
$$
$$
z=x+i y
$$
Given, $\left|\frac{z-3}{z-2 i}\right|=2$
$$
\begin{aligned}
& |(x-3)+i y|=2|x+(y-2) i| \\
& (x-3)^2+y^2=4\left(x^2+(y-2)^2\right. \\
& =x^2-6 x+9+y^2=4 x^2+4 y^2-16 y+16 \\
\Rightarrow & 3 x^2+3 y^2+6 x-16 y+7=0 \\
\Rightarrow & x^2+y^2+2 x-\frac{16}{3} y+\frac{7}{3}=0 \\
\therefore & \text { Centre of circle }\left(-1, \frac{8}{3}\right)
\end{aligned}
$$
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