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Q.

$If | z - 3 + i | = 4, then the locus of z = x + iy is$

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a

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

b

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

c

$x^{2} + y^{2} - 3 x + y - 6 = 0$

d

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

answer is A.

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Detailed Solution

$Given, | z - 3 + i | = 4 \begin{array}{l} \Rightarrow | z + iy - 3 + i | = 4 \\ \Rightarrow | x + iy - 3 + i | = 4 \\ \Rightarrow (x - 3)^{2} + (y + 1)^{2} = 16 \\ \Rightarrow x^{2} - 6 x + 9 + y^{2} + 2 y + 1 = 16 \\ \Rightarrow x^{2} + y^{2} - 6 x + 2 y - 6 = 0 \end{array}$

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