Find the intervals of the values of `a` for which the line `y+x=0` bisects two chords drawn from the point `((1+sqrt(2)a)/2,(1-sqrt(2)a)/2)` to the circle `2x^2+2y^2-(1+sqrt(2)a)x-(1-sqrt(2)a)y=0`

Answer

To solve the problem, we need to find the intervals of the values of \( a \) for which the line \( y + x = 0 \) bisects two chords drawn from the point \( P\left(\frac{1 + \sqrt{2}a}{2}, \frac{1 - \sqrt{2}a}{2}\right) \) to the circle given by the equation \( 2x^2 + 2y^2 - (1 + \sqrt{2}a)x - (1 - \sqrt{2}a)y = 0 \).

### Step 1: Rewrite the Circle Equation
First, we can rewrite the circle equation in standard form. Dividing the entire equation by 2 gives us:
\[ x^2 + y^2 - \frac{1 + \sqrt{2}a}{2}x - \frac{1 - \sqrt{2}a}{2}y = 0 \]
This can be rearranged to:
\[ x^2 + y^2 - \frac{1 + \sqrt{2}a}{2}x - \frac{1 - \sqrt{2}a}{2}y = 0 \]

### Step 2: Identify the Center and Radius
The center \( (h, k) \) of the circle can be found using the formula:
\[ h = \frac{1 + \sqrt{2}a}{4}, \quad k = \frac{1 - \sqrt{2}a}{4} \]
To find the radius \( r \), we complete the square for both \( x \) and \( y \) terms. The radius can be determined from the constant term after rearranging.

### Step 3: Use the Chord Bisector Condition
The line \( y + x = 0 \) bisects the chords if the point \( P \) lies on the line perpendicular to the chord at its midpoint. The midpoint of the chord can be expressed in terms of the parameters of the circle and the point \( P \).

### Step 4: Set Up the Quadratic Equation
Using the chord length formula and the condition that the line bisects the chords, we can derive a quadratic equation in terms of \( \alpha \) (the parameter along the line). The general form of the quadratic equation will be:
\[ A\alpha^2 + B\alpha + C = 0 \]
where \( A, B, C \) are expressions involving \( a \).

### Step 5: Find the Discriminant
For the line to bisect the chords, the quadratic must have real roots, which means the discriminant must be greater than zero:
\[ D = B^2 - 4AC > 0 \]

### Step 6: Solve the Inequality
We will solve the inequality \( D > 0 \) to find the intervals of \( a \). This will involve simplifying the expression for \( D \) and determining the values of \( a \) that satisfy the inequality.

### Step 7: Analyze the Results
After solving the inequality, we will find the critical points and test intervals to determine where the discriminant is positive.

### Final Result
The intervals of \( a \) for which the line \( y + x = 0 \) bisects the chords drawn from the point to the circle are:
\[ a \in (-\infty, -2) \cup (2, \infty) \]