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

4

Explanation

Let `B(h,0)` be the midpoint of the chord drawn from point A(p,q) .
Also, the center is `C(p//2,q//2)`.
Then, we have `BC_|_ AB`. Therfore,
`((q//2)-2)/((p//2)-h)((q-0)/(p-h))= -1`
`:. ((q)/(p-2h))((q-0)/(p-h))= -1`
`:. 2h^(2)-3ph+p^(2)+q^(2)=0`
SInce two such chords exist, the above equaton must have two distinct real roots, i.e.,
Discriminant `gt0`
`:. 9 p^(2)-8(p^(2)+q^(2)) gt0`
or `p^(2) gt 8q^(2)`