The set of all values of a ^ 2 for which the line x + y = 0 bisects...

Given,

The line $y = - x$ bisect the chord $PQ$

If $\left(\lambda, - \lambda\right)$ is the mid-point of the chord $PQ$ then,

$Q ( 2\lambda - \alpha, - 2\lambda - \beta )$

Given circle $2{x}^{2} + 2{y}^{2} - ( 1 + a ) x - ( 1 - a ) y = 0$

Or, ${x}^{2} + {y}^{2} - \left(\dfrac{1 + a}{2}\right)x - \left(\dfrac{1 - a}{2}\right)y = 0$

Let $\dfrac{1 + a}{2} = \alpha, \dfrac{1 - a}{2} = \beta$

The circle equation will be,

${x}^{2} + {y}^{2} - \alphax - \betay = 0$

Point $Q$ will lie on the circle,

$( 2\lambda - \alpha{)}^{2} + ( - 2\lambda - \beta{)}^{2} - \alpha ( 2\lambda - \alpha ) - \beta ( - 2\lambda - \beta ) = 0$

$⇒ 4{\lambda}^{2} + 3\lambda ( \beta - \alpha ) + \left({\alpha}^{2} + {\beta}^{2}\right) = 0$

So, quadratic in $\lambda$ should have $D > 0$

$⇒ 9 ( \beta - \alpha{)}^{2} - 4 ( 4 ) \left({\alpha}^{2} + {\beta}^{2}\right) > 0$

Put the value of $\alpha and \beta$ we get,

$⇒ 9{a}^{2} - 16\left(\dfrac{1 + {a}^{2}}{2}\right) > 0$

$⇒ {a}^{2} > 8$