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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 origin upon any chord which subtends a right angle at the origin.

A
2(x2+y2)+2(9x+fy)+c=0
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B
x2+y2+2(9x+fy)+c=0
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C
2x2+2y2+9xfy+c=0
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D
None of these
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Solution

The correct option is D None of these
Let, P(h,k) be the foot of perpendicular drawn from origin O(0,0) on the chord AB of the given circle such that the chord AB subtends a right angle at the origin.
The equation of chord AB is
yk=hk(xh)hx+ky=h2+k2
The combined equation of OA and OB is homogeneous equation of the second degree obtained by the help of the given circle and the chord AB and is given by
x2+y2(2gx+2fy)(hx+kyh2+k2)+c(hx+kyh2+k2)2=0
Since, the lines OA and OB are at right angle.
Therefore coefficient of x2+ coefficient of y2=0
⎪ ⎪ ⎪⎪ ⎪ ⎪1+2ghh2+k2+ch2(h2+k2)2⎪ ⎪ ⎪⎪ ⎪ ⎪+⎪ ⎪ ⎪⎪ ⎪ ⎪1+2fkh2+k2+ch2(h2+k2)2⎪ ⎪ ⎪⎪ ⎪ ⎪=02(h2+k2)+2(gh+fk)+c=0
h2+k2+gh+fk+c2=0
Therefore required equation of locus is
x2+y2+gh+fk+c2=0

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