If a circle drawn by assuming a chord parallel to the transverse...

Let, the end points of the diameter of circle are $\left(asec \theta, btan \theta\right)$ & \left(- asec \theta,btan \theta\right)$<br />$⇒$Equation of circle is$\left(x - asec \theta\right)\left(x + asec \theta\right) + {\left(y - btan \theta\right)}^{2} = 0$<br />$⇒ x^{2} - a^{2}\sec^{2} \theta + y^{2} + b^{2}\tan^{2} \theta - 2btan \thetay = 0$<br />Because \left(2,0\right)$ satisfies above equation,
hence,~$4 - a^{2}\left(1 + \tan^{2} \theta\right) + 0 + b^{2}\tan^{2} \theta = 0$
$⇒ \left(4 - a^{2}\right) + \left(b^{2} - a^{2}\right)\tan^{2} \theta = 0$
Which is always true if $\left|a\right| = 2$ & \left|b\right| = 2