The set of all values of a^2 for which the line x+y=0 bisects two distinct chords drawn fr…

Question

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

Accepted Answer

Circle: $x^2+y^2-\dfrac{(1+a)}2x-\dfrac{(1-a)}2y=0$, centre $(h,k)=\left(\dfrac{1+a}4,\dfrac{1-a}4\right)$; note $P=(2h,2k)$. A chord bisected at $(\lambda,-\lambda)$ (on $x+y=0$) has equation $T=S_1$; requiring this chord to pass through $P$ gives, after substitution, the quadratic in $\lambda$: $$2\lambda^2+3(k-h)\lambda+(hx_P+ky_P)=0,\quad\text{i.e.}\quad 2\lambda^2+3(k-h)\lambda+2(h^2+k^2)=0.$$ Two distinct real chords require discriminant $>0$: $$9(k-h)^2-16(h^2+k^2)>0\ \Longrightarrow\ 7h^2+7k^2+18hk<0.$$ Substituting $h=\dfrac{1+a}4,\ k=\dfrac{1-a}4$ and simplifying: $$\frac{7(2+2a^2)}{16}+\frac{18(1-a^2)}{16}<0\ \Longrightarrow\ 32-4a^2<0\ \Longrightarrow\ a^2>8.$$ So $a^2\in(8,\infty)$.

Solutions · 1