1. Use the theorem for chords bisected by a line

For a circle, if a line bisects all chords drawn through a point $P$, then that line is the polar of $P$ with respect to the circle.
Conversely, if a line passes through the center, then chords through points on that line are bisected trivially.

Here, we need the line $x+y=0$ to bisect two distinct chords drawn from the point $P\left(\frac{1+a}{2},\frac{1-a}{2}\right)$ on the circle $2x^2+2y^2-(1+a)x-(1-a)y=0.$

A standard result: if from a point $P$ on/inside/outside a circle, chords through $P$ are bisected by a fixed line $L$, then the midpoint locus condition implies that either:

• $L$ is the polar of $P$, giving exactly one chord generally, or
• $P$ lies on the diameter perpendicular relation giving two distinct chords iff $P$ is inside the circle and the line is not tangent in the corresponding transformed setup.

A cleaner analytic approach is better.

2. Write the circle in standard form

Divide by $2$: $x^2+y^2-\frac{1+a}{2}x-\frac{1-a}{2}y=0.$

Comparing with $x^2+y^2+2gx+2fy+c=0,$ we get $2g=-\frac{1+a}{2},\qquad 2f=-\frac{1-a}{2},\qquad c=0.$ So $g=-\frac{1+a}{4},\qquad f=-\frac{1-a}{4}.$

Hence center is $C=(-g,-f)=\left(\frac{1+a}{4},\frac{1-a}{4}\right).$

Radius: $r^2=g^2+f^2-c=\frac{(1+a)^2+(1-a)^2}{16}=\frac{2+2a^2}{16}=\frac{1+a^2}{8}.$ So $r=\sqrt{\frac{1+a^2}{8}}.$

3. Check that $P$ lies on the circle

Given $P\left(\frac{1+a}{2},\frac{1-a}{2}\right)=\left(2\cdot \frac{1+a}{4},2\cdot \frac{1-a}{4}\right)=2C.$ So $P$ is the point obtained by doubling the center coordinates from origin. Indeed substituting confirms it lies on the circle.

Also note $C$ is the midpoint of $O(0,0)$ and $P$. Therefore the circle has diameter $OP$.

4. Condition for line $x+y=0$ to bisect chords through $P$

Take any chord through $P$. Let its other endpoint be $Q$. If the midpoint $M$ of $PQ$ lies on the line $x+y=0$, then reflecting the chord condition analytically gives that $Q$ must satisfy the circle equation and $\frac{x_P+x_Q}{2}+\frac{y_P+y_Q}{2}=0.$ Since $x_P+y_P=\frac{1+a}{2}+\frac{1-a}{2}=1,$ this becomes $x_Q+y_Q=-1.$

So the other endpoint $Q$ must lie on the line $x+y=-1.$ Thus, the required chords correspond exactly to intersections of the circle with the line $x+y=-1.$

For the line $x+y=0$ to bisect two distinct chords through $P$, the line $x+y=-1$ must intersect the circle in two distinct points.

So we need the distance from center $C$ to the line $x+y=-1$ to be strictly less than the radius.

5. Compute distance from center to the line $x+y+1=0$

Center: $C\left(\frac{1+a}{4},\frac{1-a}{4}\right).$ Then $x_C+y_C=\frac{1+a}{4}+\frac{1-a}{4}=\frac12.$

Distance from $C$ to $x+y+1=0$ is $d=\frac{\left|\frac12+1\right|}{\sqrt{1^2+1^2}}=\frac{3/2}{\sqrt2}=\frac{3}{2\sqrt2}.$ Thus $d^2=\frac{9}{8}.$

6. Impose two distinct intersections

We need $d<r$ which gives $\frac{9}{8}<\frac{1+a^2}{8}.$ So $9<1+a^2$ $a^2>8.$

If $a^2=8$, the line is tangent, giving only one chord (degenerate case), not two distinct chords. Hence $a^2\in(8,\infty).$

7. Match with options

This is exactly Option D.

$\boxed{(8,\infty)}$