The set of all values of $a^2$ for which the line $x+y=0$ bisects two distinct chords drawn from a point $\mathrm{P}\left(\frac{1+\mathrm{a}}{2},\frac{1-\mathrm{a}}{2}\right)$ on the circle $2{\mathrm{x}}^2+2{\mathrm{y}}^2-(1+\mathrm{a})\mathrm{x}-(1-\mathrm{a})\mathrm{y}=0$ is equal to :

${\mathrm{x}}^2+{\mathrm{y}}^2-\left(\frac{1+\mathrm{a}}{2}\right)\mathrm{x}-\left(\frac{1-\mathrm{a}}{2}\right)\mathrm{y}=0 ...\left(1\right)$
Clearly circle (1) passes through (0,0)
$\mathrm{O}(0,0),\mathrm{P}\left(\frac{1+\mathrm{a}}{2},\frac{1-\mathrm{a}}{2}\right)$are ends of diameter
$\mathrm{L}(\mathrm{\alpha },-\mathrm{\alpha })$be the midpoint of chord PQ
$\frac{\mathrm{P}+\mathrm{Q}}{2}=\mathrm{L}\Rightarrow \mathrm{Q}=2\mathrm{L}-\mathrm{P}$
$\begin{array}{l}2(\mathrm{\alpha },-\mathrm{\alpha })-\left(\frac{1+\mathrm{a}}{2},\frac{1-\mathrm{a}}{2}\right)\\ \mathrm{Q}\left(2\mathrm{\alpha }-\left(\frac{1+\mathrm{a}}{2}\right),-2\mathrm{\alpha }-\left(\frac{1-\mathrm{a}}{2}\right)\right)\end{array}$
Q lies on circle ….(1) simplifying
We get $8{\mathrm{\alpha }}^2-6\mathrm{a\alpha }+\left(1+{\mathrm{a}}^2\right)=0$ is a Q.E in ${ }^{\text{'}}\mathrm{\alpha }\text{ ' but }\mathrm{\alpha }\in \mathrm{ℝ},\mathrm{\alpha }$ is 
$\begin{array}{l}{\mathrm{b}}^2-4\mathrm{ac}>0\\ 36{\mathrm{a}}^2-4\cdot 8\cdot \left(1+{\mathrm{a}}^2\right)>0\\ 9{\mathrm{a}}^2-8-8{\mathrm{a}}^2>0\Rightarrow {\mathrm{a}}^2-8>0\Rightarrow {\mathrm{a}}^2>8\\ {\mathrm{a}}^2\in (8,\mathrm{\infty })\end{array}$