Question

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

Answer 1

Let the given point be $(p,\overline{p})=(\frac{1+\sqrt{2}a}{2},\frac{1-\sqrt{2}a}{2})$ and the equation of the circle becomes
$x^2+y^2-px-\overline{p}y=0$
Since the chord is bisected by the line $x+y=0$, its mid-point can be chosen as $(k,-k)$ on this line. Hence the equation of the chord by $T=S_1$ is
$kx-ky-\frac{p}{2}(x+k)-\frac{\overline{p}}{2}(y-k)=k^2+k^2-pk-\overline{p}k$
It passes through $A(p,\overline{p})$.
$\therefore kp=k\overline{p}-\frac{p}{2}(p+k)-\frac{\overline{p}}{2}(\overline{p}-k)=2k^2-pk+\overline{p}k$
or $3k(p-\overline{p})=4k^2+(p^2-{\overline{p}}^2).....\left(1\right)$
Put $p-\overline{p}=a\sqrt{2}$,
$p^2+{\overline{p}}^2=2.\frac{1+2a^2)}{4}=\frac{1+2a^2}{2}.....\left(2\right)$
Hence from $\left(1\right)$ by the help of $\left(2\right)$, we get
$4k^2-3\sqrt{2}ak+\frac{1}{2}(1+2a^2)=0.....\left(3\right)$
Since there are two chords which are bisected by $x+y=0$, we must have two real values of $k$ from $\left(3\right)$
$△>0$ or $18a^2-8(1+2a^2)>0$
or $a^2-4>0$ or $(a+2)(a-2)>0$
$\therefore a<-2$ or $>2$

$\therefore aϵ(-\infty ,-2)\cup (2,\infty )$