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 8 q^(2)`
D. `p^(2) lt 8q^(2)`

Answer

D

Explanation

Let `(x_(1), 0)` be the mid-point of the chords drawn from the point (p, q). Then, the equation of the chord bisected at `(x_(1),0)` is
`x_(1)^(2)-px_(1)-x x_(1) + 0y - p ((x+x_(1))/(2))-q ((y+0)/(2))` [ Using: X'=T]
`rArr 2x_(1)^(2)-2px_(1)=2x x_(1)-p(x +x_(1))-qy`
This passes through (p, q).
`:. 2x_(1)^(2)-2p x_(1)=2px_(1)-p(p+x_(1))-q^(2)`
`rArr 2x_(1)^(2)-3px_(1)+(p^(2)+q^(2))=0`
Now, `x_(1)` is real.
`:. 9p^(2)-8(p^(2)+q^(2)) gt 0 rArr p^(2)-8q^(2) gt 0 rArr p^(2) gt 8q^(2)`