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Q.

Chords of the circle x 2 + y 2 +2gx+2fy+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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a

x 2 + y 2 +gx+fy+c=0  

b

2 x 2 + y 2 +gx+fy +c=0  

c

2 x 2 + y 2 +gx+fy+c=0  

d

x 2 + y 2 +2(gx+fy+c)=0   

answer is C.

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Detailed Solution

detailed_solution_thumbnail

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 1 2 + y 1 2 ⇒ x x 1 +y y 1 x 1 2 + y 1 2 =1  

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

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