Multiple choice

Let $S=x^{2}+y^{2}+2gx+2fy+c=0$ be a given circle . Then the locus of the foot of the perpendicular drawn from the origin upon any chords of $S$ which subtends right angle at the origin is:

  1. $x^{2}+y^{2}+gx+fy+c/2=0$
  2. $x^{2}+y^{2}=g$
  3. $x^{2}+y^{2}=f$
  4. $x^{2}+y^{2}+g=0$
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A Correct answer
Explanation

If a chord of x^2 + y^2 + 2gx + 2fy + c = 0 subtends a right angle at the origin, the locus of the foot of the perpendicular from the origin to the chord is x^2 + y^2 + gx + fy + c/2 = 0. This is a standard result derived from the equation of the chord of contact and the condition of perpendicularity.

AI explanation

Let the chord joining points P and Q subtend a right angle at the origin, meaning the equation of the pair of lines OP and OQ represents a homogeneous second-degree equation. If (h, k) is the foot of the perpendicular from the origin to the chord PQ, then substituting the line equation into the circle's equation allows us to find the locus of (h, k). By using the standard geometric property for a chord subtending a right angle at the center, the locus is derived as half the given circle's equation, yielding the result x^2 + y^2 + gx + fy + c/2 = 0.

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