IF TWO DISTINCT CHORDS, DRAWN FROM THE POINT (P,Q) TO THE CIRCLE   x² + y² = px + qy

( pq not equal to 0) ARE BISECTED BY X - AXIS, THEN FIND RELATION BETWEEN P AND Q

The ans is P²>8Q².Posting the soln in next post,...

Since the chords drawn from (p,q) are bisected by X-axis,The midpoint of these chords is of the form (a,0) and hence the other extremity of the chord becomes (2a-p,-q).This lies on the circle iff (2a-p)²+q²=p(2a-p)-q² iff 2a²-3ap+p²+q²=0

Now,this quadratic shud have two distinct roots as two distinct are bisected by x-axis,hance the discriminant 9p²-8(p²+q²)>0p²>8q²

There is one more method to do this question.

that is by using the equation of the chord whose mid pt. is given  S₁=T

let the mid pt be (a,0)

Therefore we get     a²-ap=ax -px/2 -ap/2 - qy/2

now since pt. (p,q) lies on chord we can satisfy the equation with it

a²-ap-ap+p²/2+ap/2+q²/2=0

2a²-4ap+p²+ap+q²=0

2a²-3ap+(p²+q²)=0

Now,this quadratic shud have two distinct roots as two distinct are bisected by x-axis,hance the discriminant D>0

9p²-8(p²+q²)>0

thus we get p²>8q²

Plz rate if the soln. was helpful.