Animated Solution for Mathematics - Circles: Find the intervals of values of $a$ for which the line $y+x=0$ bisects two chords drawn from a point $(\frac{1+\sqrt{2}a}{2},\frac{1-\sqrt{2}a}{2})$ to the circle $2x^2+2y^2-(1+\sqrt{2}a)x-(1-\sqrt{2}a)y=0$.

Imagine a circle defined by the equation:

We are given a line $y+x=0$ and a point $P$ located on the circle's boundary. We seek the values of $a$ such that the line bisects exactly two chords originating from $P$.

The key to unlocking this problem lies in the definition of a bisected chord. If a line bisects a chord, the midpoint of that chord must lie on the line.

Since our line is $y+x=0$, any midpoint $M$ must satisfy $y=-x$. Let the coordinates of this midpoint be $M(h,-h)$.

We invoke the $T=S_1$ theorem. For a circle $S=0$, the equation of a chord with a known midpoint $(h,k)$ is given by $T=S_1$.

To simplify the algebra, let:

The circle equation simplifies to $x^2+y^2-px-qy=0$. Applying $T=S_1$ at $M(h,-h)$, we obtain:

We know the chord must pass through $P(p,q)$. By substituting $x=p$ and $y=q$ into our chord equation, we transform the geometry into a quadratic equation in $h$:

This quadratic is the heart of the problem, as it determines the possible positions of the midpoints. For the line to bisect two distinct chords, this quadratic must yield two distinct real roots for $h$.

This requires the discriminant $D>0$.

We calculate the necessary components:

Substituting these into the discriminant condition $D=[-3(p-q){]}^2-4(4)(p^2+q^2)>0$, we get:

This simplifies to:

Thus, the values of $a$ are $a\in (-\infty ,-2)\cup (2,\infty )$. We have successfully navigated the complexity, turning a daunting geometric puzzle into a clear, elegant solution.