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Question

If Sx2+y2+2gx+2fy+c=0 is a given circle, then the locus of the foot of the perpendicular drawn from origin upon any chord of S which subtends a right angle at the origin, is

A
2(x2+y2)+2gx+2fy+c=0
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B
2(x2+y2)+2gx+2fyc=0
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C
(x2+y2)+gx+fy+c=0
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D
None of these
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Solution

The correct option is A 2(x2+y2)+2gx+2fy+c=0
Let P be the foot of from origin O on any chord of the circle S whose coordinates are (α,β).
Then, the slope of OP is βα and thus the slope of chord is αβ and its equation passing through (α,β) is yβ=αβ(xα)βyβ2=αx+α2
αx+βy=α2+β2 ...(1)
Now, homogenizing the equation of the given circle
x2+y2+2gx+2fy+c=0
with the help of (1), we get
x2+y2(2gx+2fy)(αx+βyα2+β2)+c(αx+βyα2+β2)2=0. ...(2)
Now, equation (2) represents a pair of straight lines pass-ing through origin.
These lines will be at right angle if sum of the coefficients of x2 and y2 is zero.
i.e., (α2+β2)2+(α2+β2)2+2gα(α2+β2)+2βf(α2+β2)+c(α2+β2)
2(α2+β2)+2gα+3fβ+c=0 ...(3)
From equation (3), the locus of P(α,β) is
2(x2+y2)+2gx+2fy+c=0
which is the required locus.

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