The equation of the circle is 2x(x–a) + y(2y–b)=0 (a $\ne$ 0, b $\ne$ 0) . The condition on a and b if two chords, each bisected by the x-axis, can be drawn to the circle from $\left(\mathrm{a},\frac{\mathrm{b}}{2}\right)$ is 

$\left(\frac{\mathrm{a}+\mathrm{\alpha }}{2},\frac{\frac{\mathrm{b}}{2}+\mathrm{\beta }}{2}\right)=\left({\mathrm{x}}_1,0\right)$
$\Rightarrow \mathrm{\alpha }=2{\mathrm{x}}_1-\mathrm{a}\Rightarrow \mathrm{\beta }=\frac{-\mathrm{b}}{2}$

(α,ß) lies in 2x² + 2y² – 2ax – by = 0

$$\begin{gathered} 2{\left(2x_1-a\right)}^2+2{\left(\frac{-b}{2}\right)}^2-2a\left(2x_1-a\right)-b\left(\frac{-b}{2}\right)=0 \\ 2\left(4{x_1}^2+a^2-4a{x}_1\right)+2\left(\frac{b^2}{4}\right)-4a{x}_1+2a^2+\frac{b^2}{2}=0 \\ 8{x_1}^2+2a^2-8a{x}_1-4a{x}_1+2a^2+b^2=0 \\ 8{x_1}^2-12a{x}_1+4a^2+b^2=0 \end{gathered}$$

the value of x₁ is real and distinct then,

$$\begin{gathered} b^2-4ac>0 \\ 144a^2-4\left(8\right)\left(4a^2+b^2\right)>0 \\ 9a^2-8a^2-2b^2>0 \\ {a}^2-2b^2>0 \end{gathered}$$