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If $| Z + 2 - 3 i | = 5$ then locus of Z = x + iy is

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$x^{2} + y^{2} + x - y = 0$

$x^{2} + y^{2} + 4 x - 6 y - 12 = 0$

$x^{2} + y^{2} + 3 x - 2 y = 0$

$3 x^{2} + 3 y^{2} - 22 x + 35 = 0$

answer is B.

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Detailed Solution

$| Z + 2 - 3 i | = 5 \Rightarrow {(x + 2)}^{2} + {(y - 3)}^{2} = 25 \Rightarrow x^{2} + y^{2} + 4 x - 6 y - 12 = 0$

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