Circle 4 Question 20

20. Let Sx2+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.

(1988,5 M)

Show Answer

Answer:

Correct Answer: 20. x2+y2+gx+fy+c2=0

Solution:

  1. 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 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+chx+kyh2+k22=0

Since, the lines OA and OB are at right angle. Coefficient of x2+ Coefficient of y2=0

1+2ghh2+k2+ch2(h2+k2)2

+1+2fkh2+k2+ck2(h2+k2)2=0

2(h2+k2)+2(gh+fk)+c=0

h2+k2+gh+fk+c2=0

Required equation of locus is

x2+y2+gx+fy+c2=0



NCERT Chapter Video Solution

Dual Pane