If $p + i q = {tan}^{- 1} (z), z = x + i y$ and $p$ is constant then locus of $z$ is

x^{2} + y^{2} + 2 x cot 2 p = 1

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cot 2 p (x^{2} + y^{2}) = 1 + x

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x^{2} + y^{2} + 2 y tan 2 p = 1

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x^{2} + y^{2} + 2 y tan 2 p = - 1

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Solution

The correct option is A $x^{2} + y^{2} + 2 x cot 2 p = 1$
Given,
$t a n^{- 1} (x + i y) = p + i q$
also, $t a n^{- 1} (x - i y) = p - i q$

SO,
$t a n^{- 1} (x + i y) + t a n^{- 1} (x - i y) = 2 p ⟹ t a n^{- 1} \frac{(x + i y) (x - i y)}{1 - (x + i y) 9 x - i y)} = 2 p$

so , $\frac{2 x}{1 - (x^{2} + y^{2})} = t a n 2 p$

or $2 x c o t 2 p = 1 - (x^{2} + y^{2})$

or $(x^{2} + y^{2}) + 2 x c o t 2 p = 1$

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