Let $x(x-a)+y(y-1)=0$ be a circle. If two chords from (a, 1) bisected by X-axis are drawn to the circle then the condition is

Given circle ${\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{ax}-\mathrm{y}=0$

Let a (A, 1),  B (x, y) be extremities of a chord bisected by x-axis then mid point of AB lies on x-axis

$\begin{array}{rcl} & \Rightarrow & \frac{\mathrm{y}+1}{2}=0\\ & \Rightarrow & \mathrm{y}+1=0\\ & \Rightarrow & \mathrm{y}=-1\end{array}$

$\therefore (\mathrm{x}, -1)$ lien on the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-\mathrm{ax}-\mathrm{y}=0$

$\begin{array}{rcl} & \Rightarrow & {\mathrm{x}}^2+1-ax+1=0\\ & \Rightarrow & {\mathrm{x}}^2-ax+2=0\end{array}$

since two chords are possible then

 ${\mathrm{x}}^2-\mathrm{ax}+2=0$ has two distinct roots 

$\Rightarrow {\mathrm{a}}^2-4\left(2\right)>0$ $( \therefore$ discriminant greater than zero $)$

$\Rightarrow {\mathrm{a}}^2>8$