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(C) Given,$\left|\frac{z-3}{z-2i}\right|=2$. Substituting $z=x+iy$,we get $|(x-3)+iy| = 2|x+i(y-2)|$. Squaring both sides,we have $(x-3)^2 + y^2 = 4(x

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$\left(-1, \frac{8}{3}\right)$

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(C) Given,$\left|\frac{z-3}{z-2i}\right|=2$. Substituting $z=x+iy$,we get $|(x-3)+iy| = 2|x+i(y-2)|$. Squaring both sides,we have $(x-3)^2 + y^2 = 4(x^2 + (y-2)^2)$. Expanding the terms,$x^2 - 6x + 9 + y^2 = 4(x^2 + y^2 - 4y + 4)$. $x^2 - 6x + 9 + y^2 = 4x^2 + 4y^2 - 16y + 16$. Rearranging the terms,$3x^2 + 3y^2 + 6x - 16y + 7 = 0$. Dividing by $3$,we get $x^2 + y^2 + 2x - \frac{16}{3}y + \frac{7}{3} = 0$. Comparing this with the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$,we have $2g = 2 \Rightarrow g = 1$ and $2f = -\frac{16}{3} \Rightarrow f = -\frac{8}{3}$. The centre of the circle is $(-g, -f) = (-1, \frac{8}{3})$.

If $z=x+iy$,then the centre of the circle $\left|\frac{z-3}{z-2i}\right|=2$ is

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