Let a circle be given by `2x(x-a)+y(2y-b)=0,(a cancel(=)0,b cancel(=)0)`. Find the condition on a anb b if two chords , each bisected by the x-axis, can be drawn to the circle from `(a,b//2)`.

Explanation

The given circle can be rewritten as
`x^(2)+y^(2)-ax-(by)/(2)=0" "...(i)`
Let one of the chord through (a, b/2) be bisected at (h, 0). Then, the equation of the chord having (h, 0) as mid-point is 0
`T=S_(1)`
`rArr h*x+0*y-(a)/(2)(x+h)-(b)/(h)(y+0)=h^(2)+0-ah-0`
`rArr (h-(a)/(2))x-(by)/(4)-(a)/(2)h=h^(2)-ah" "...(ii)`
It passes through `(a, b//2)`,then
`(h-(a)/(2))a-(b)/(4)*(b)/(2)-(a)/(2)=h^(2)-ah`
According to the given condition, Eq. (iii) must have tow distinct real roots. This is possible, if the discriminant of Eq. (iii) is greater than 0.
i.e. `(9)/(4)a^(2)-4((a^(2))/(2)+b^(2)/8)gt0rArr (a^(2))/(4)-(b^(2))/(2)gt0`
`rArr a^(2)gt2b^(2)`