If z=x+iy, then the centre of the circle z−3z−2i=2 is

z−3z−2i=2⇒|(x−3)+iy|=2|x+(y−2)i|⇒(x−3)2+y2=4x2+(y−2)2⇒x2+y2−6x+9=4(x2+y2−4y+4)⇒3x2+3y2+6x−16y+7=0∴Centre -1,8/3

If z=x+iy, then the centre of the circle z−3z−2i=2 is

z−3z−2i=2⇒|(x−3)+iy|=2|x+(y−2)i|⇒(x−3)2+y2=4x2+(y−2)2⇒x2+y2−6x+9=4(x2+y2−4y+4)⇒3x2+3y2+6x−16y+7=0∴Centre -1,8/3

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

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$(1, - 8 / 3)$

$(- 1, 8 / 3)$

$(- 1, - 8 / 3)$

$(1, 8 / 3)$

answer is C.

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$|\frac{z - 3}{z - 2 i}| = 2 \Rightarrow | (x - 3) + i y | = 2 | x + (y - 2) i | \Rightarrow {(x - 3)}^{2} + y^{2} = 4 [x^{2} + {(y - 2)}^{2}]$

$\Rightarrow x^{2} + y^{2} - 6 x + 9 = 4 (x^{2} + y^{2} - 4 y + 4) \Rightarrow 3 x^{2} + 3 y^{2} + 6 x - 16 y + 7 = 0$

$∴$ Centre $(- 1, 8 / 3)$

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