$x ^2+ y ^2-\frac{(1+ a ) x }{2}-\frac{(1- a ) y }{2}=0$

$\text { Centre }\left(\frac{1+ a }{4}, \frac{1- a }{4}\right) \Rightarrow( h , k )$

$P \left(\frac{1+ a }{2}, \frac{1- a }{2}\right) \Rightarrow(2 h , 2 k )$

Equation of chord $\Rightarrow T=S_1$

$\Rightarrow( x – y ) \lambda-\frac{2 h ( x +\lambda)}{2}-\frac{(2 k )( y -\lambda)}{2}$

$=2 \lambda^2-2 h (\lambda)+2 k \lambda$

Now, $\lambda(2 h , 2 k )$ satisfies the chord

$\therefore(2 h -2 k ) \lambda- h ( x +\lambda)- k ( y -\lambda)$

$\Rightarrow 2 \lambda^2+4 k \lambda-4 h \lambda+ h \lambda- k \lambda+ hx + ky =0$

$\Rightarrow 2 \lambda^2+\lambda(3 k -3 h )+ ky + hx =0$

$\Rightarrow D > 0$

$\Rightarrow 9( k – h )^2-8( ky + hx ) > 0$

$\Rightarrow 9( k – h )^2-8\left(2 k ^2+2 h ^2\right) > 0$

$\Rightarrow-7 k ^2-7 h ^2-18 kh > 0$

$\Rightarrow 7 k ^2+7 h ^2+18 kh < 0$

$\Rightarrow 7\left(\frac{1- a }{4}\right)^2+7\left(\frac{1+ a }{4}\right)^2+18\left(\frac{1- a ^2}{16}\right) < 0$

$\Rightarrow 7\left[\frac{2\left(1+ a ^2\right)}{16}\right]+\frac{18\left(1- a ^2\right)}{16} < 0, \quad a ^2= t$

$\Rightarrow \frac{7}{8}(1+ t )+\frac{18(1- t )}{16} < 0$

$\Rightarrow \frac{14+14 t +18-18 t }{16} < 0$

$\Rightarrow 4 t > 32$

$t > 8 \quad a ^2 > 8$