If two distinct chords, drawn from the point (p, q) on the circle `x^2+y^2=p x+q y` (where `p q!=q)` are bisected by the x-axis, then `p^2=q^2` (b) `p^2=8q^2` `p^2<8q^2` (d) `p^2>8q^2`

Options

A. `p^(2)=q^(2)`
B. `p^(2)=8q^(2)`
C. `p^(2)lt 8q^(2)`
D. `p^(2)gt 8q^(2)`

Answer

D

Explanation

Chord is bisected by x- axis , so that its mid point is (h,O) Hence by `T=s_(1)`
its equation is
`bx +0-(P)/(2)(y+0)= h^(2)-ph`
it passes through the point (p,q)
`therefore 2 h^(2)- 3ph+(p^(2)+q^(2))=0" ".....(1) `
it is a quadratic equation in h Since two distinct chords are drawn the roots of (1) will be real and distinct
`therefore Dgt0implies9p^(2)-8(p^(2)+q^(2))gt 0`
or ` p^(2) gt 8 q^(2)`