If two distinct chords drawn from the point
(a,b) on the circle x^2+y^2=ax+by
when ab not equal to zero
are bisected by the x-axis
prove that a^2=8b^2
I'd start with this:
The circle's equation is
(x - a/2)^2 + (y - b/2)^2 = (a^2+b^2)/4
If a chord intersects the x-axis at x=c, then the slope of the chord is
(b)/(a-c)
so, the equation of the chord is
y = b/(a-c) (x-c)
and the length of the chord is
2sqrt((a-c)^2 + b^2)
See where that takes you