Consider the circle C : x2+y2-6x-8y-11=0. Let a variable chord AB of the circle C subtend a right angle at the origin. If the locus of the foot of the perpendicular drawn from the origin on the chord AB is the circle x2+y2- x- y- =0, then + +2 is equal to .
Answer
18
Solution
[figure]
Let M(h, k) be the foot of the perpendicular from O on AB OM AB
AB: hx+ky=h2+k2=p (say)
Homogenising C with hx+kyp=1 gives the pair of lines OA, OB:
x2+y2-(6x+8y) hx+kyp-11 (hx+ky)2p2=0
=90 coefficient of x2 + coefficient of y2=0
(1- 6hp- 11h2p2 )+ (1- 8kp- 11k2p2 )=0
2- 6h+8kp- 11p=0 ( h2+k2=p )
2 (h2+k2 )-6h-8k-11=0 h2+k2-3h-4k- 112=0
=3, =4, = 112 + +2 =3+4+11=18
Answer: 18