Let `S-=x^2+y^2+2gx+2f y+c=` 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.

Answer

`x^(2)+y^(2)+gx+fy+(c)/(2)=0`

Explanation

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 subtenda a right angle at the origin.

The equation of chord AB is
`y-k=-(h)/(k)(x-h)rArr hx+ky=h^(2)+k^(2)`
The combined equaition of OA and OB is homogenous equation of second degree obtained by the help of the given circle and the chord AB and is given by,
`x^(2)+y^(2)+(2gx+2fy)((hx+ky)/(h^(2)+k^(2)))+c((hx+ky)/(h^(2)+k^(2)))^(2)=0`
Since, the lines OA and OB are at right angle.
`therefore` Coefficient of `x^(2)` + coefficient of `y^(2)` = 0
`rArr {1+(2gh)/(h^(2)+k^(2))+(ch^(2))/((h^(2)+k^(2))^(2))}`
`+ {1+(2fk)/(h^(2)+k^(2))+(ck^(2))/((h^(2)+k^(2))^(2))}=0`
`rArr 2(h^(2)+k^(2))+2(gh+fk)+c=0`
`rArr h^(2)+k^(2)+gh+fk+(c)/(2)=0`
`therefore` Required equation of locus is
`x^(2)+y^(2)+gx+fy+(c)/(2)=0`