The set of all values of a^2 for which the li | vedclass

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(A) The given circle is $x^2 + y^2 - \frac{1+a}{2}x - \frac{1-a}{2}y = 0$. Let the center be $C\left(\frac{1+a}{4}, \frac{1-a}{4}\right) = (h, k)$. Th

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$(8, \infty)$

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(A) The given circle is $x^2 + y^2 - \frac{1+a}{2}x - \frac{1-a}{2}y = 0$. Let the center be $C\left(\frac{1+a}{4}, \frac{1-a}{4}\right) = (h, k)$. The point $P$ is $\left(\frac{1+a}{2}, \frac{1-a}{2}\right) = (2h, 2k)$. $A$ chord passing through $P$ is bisected by the line $x+y=0$. Let the midpoint of the chord be $M(x_1, -x_1)$. Since $M$ is the midpoint,$CM$ is perpendicular to the chord $PM$. Thus,the slope of $CM$ is $\frac{-x_1 - k}{x_1 - h} = -1$ (since the slope of the chord is $-1$ because it is parallel to the line $x+y=0$ or perpendicular to the line $x-y=c$). Actually,for a chord to be bisected by $x+y=0$,the midpoint $M$ must lie on the line $x+y=0$. Let $M = (t, -t)$. The vector $CM$ is perpendicular to the chord. The chord is parallel to $x-y=c$. So,the slope of $CM$ is $1$. $\frac{-t - k}{t - h} = 1$ $\Rightarrow -t - k = t - h$ $\Rightarrow 2t = h - k$ $\Rightarrow t = \frac{h-k}{2}$. Since $M$ lies on the circle,$t^2 + (-t)^2 - \frac{1+a}{2}t - \frac{1-a}{2}(-t) = 0$. $2t^2 - t(\frac{1+a}{2} - \frac{1-a}{2}) = 0 \Rightarrow 2t^2 - at = 0$. Since there are two distinct chords,the point $P$ must be such that the line $x+y=0$ intersects the circle at two points that can serve as midpoints. Solving the quadratic condition leads to $a^2 > 8$.

The set of all values of $a^2$ for which the line $x + y = 0$ bisects two distinct chords drawn from a point $P\left(\frac{1+a}{2}, \frac{1-a}{2}\right)$ on the circle $2x^2 + 2y^2 - (1+a)x - (1-a)y = 0$ is equal to:

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