Circle has a quadratic equation

Question:
If two distinct chords from (p,q) on the circle
!!(x^2 + y^2)!! !!=!! !!(px+qy)!! are bisected by !!x-!!axis.Then !!a)p^2!! !!=!! !!q^2!! !!b)p^2!! !!=!! !!8q^2!! !!c)p^2!! !!>!! !!8q^2!! !!d)p^2!! !!

Solution:

Since the x axis bisects the chord, so if !!(a,b)!! is the other end of chord.

Then !!((a+p)/2,(b+q)/2)!! lies on the x axis. Hence !!b+q=0!!, !!b=-q!!.

So the pair of distinct chords are passing through !!(p,q)!! and !!(a,-q)!!.

Putting !!(a,-q)!! in circle's equation we get a quadratic equation
!!(a^2 - pa + 2q^2)!! !!=!! !!0!! For pair of distinct roots:
!!D!! !!>!! !!0!! That is
!!(p^2 - 8q^2)!! !!>!! !!0!! !!p^2!! !!>!! !!8q^2!!

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