Let S ≡ x2 + y2 + 2gx + 2fy + c = 0 Be A Given Circle. Find The Locus Of The Foot Of The Perpendicular Drawn From The Origin Upon Any Chord Of S which Subtends A Right Angle At The Origin.

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Question

Let S  x2 + y2 + 2gx + 2fy + c = 0 be a given circle. find the locus of the foot of the perpendicular drawn from the origin upon any chord of S which subtends a right angle at the origin.

Solution

Correct option is

Equation of the given circle is 

                   S = x2 + y2 + 2gx + 2fy + c = 0              …(1)

Any chord of the circle be 

                   lx + my = 1                                            …(2)  

Chord (2) subtends a right angle at the origin  So making equation (1) homogeneous with the help of (2), we get  

          (x2 + y2) + (2gx + 2fy) (lx + my) + c(lx + my)2 = 0  

          x2 + y2 + 2glx2 + 2gmxy + 2flxy + 2fmy2 + cl2x2 + cm2y2 + 2clmxy = 0  

             

Apply the condition of perpendicularity 

We get

                   

  

Any line through origin ⊥ to (1) is  

                                                                 mx – ly = 0            …(4)  

Both (2) and (4) give the foot of perpendicular whose locus is obtained by eliminating the variable lm between (2), (3) and (4).

On solving (2) & (4)

We get,      

                         

Putting in (2) we get

             

                

                         

  

Putting the value of l & m in  (3) we get  

             

  

 

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