Let P be any moving point on the circle $x^{2} + y^{2} - 2 x = 1$ AB be the chord of contact of this point w.r.t. the circle $x^{2} + y^{2} - 2 x = 0$ The locus of the circumcenter of triangle CAB (C being the center of the circle) is

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

$2 x^{2} + 2 y^{2} - 4 x + 3 = 0$

$2 x^{2} + 2 y^{2} - 4 x + 1 = 0$

$x^{2} + y^{2} - 4 x + 1 = 0$

answer is A.

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Detailed Solution

Let P be $(1 + \sqrt{2} \cos θ, \sqrt{2} \sin θ) and C be (1, 0)$
The circumcenter of triangle ABC is the midpoint of PC.
Therefore,
$2 h = 1 + \sqrt{2} \cos θ + 1 and 2 k = \sqrt{2} \sin θ$
or $[2 (h - 1)]^{2} + (2 k)^{2} = 2$
or $2 (h - 1)^{2} + k^{2} - 1 = 0$
or $2 x^{2} + 2 y^{2} - 4 x + 1 = 0$