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Chords of the circle $x 2 + y 2 + 2 g x + 2 f y + c = 0$ subtends a right angle at the origin.
The locus of the feet of the perpendiculars from the origin to these chords is

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x 2 + y 2 + g x + f y + c = 0

2 x 2 + y 2 + g x + f y + c = 0

2 x 2 + y 2 + g x + f y + c = 0

x 2 + y 2 + 2 (g x + f y + c) = 0

answer is C.

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Detailed Solution

Let $p (x, y)$ be the foot of the perpendicular then p is mid point of the chord

equation of chord is $S 1 = S 11$

$⇒ x x 1 + y y 1 = x 12 + y 12 ⇒ x x 1 + y y 1 x 12 + y 12 = 1$

Homogenize with the circle and coefficient of $x 2 +$ coefficient of $y 2 = 0$

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