Question

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 :

• A. $(8,\infty )$
• B. $(0,4]$
• C. $(4,\infty )$
• D. $(2,12]$

Answer 1

Answer: (A)

$x^2+y^2-\frac{(1+a)x}{2}-\frac{(1-a)y}{2}=0$
$\text{ Centre }\left(\frac{1+a}{4},\frac{1-a}{4}\right)\Rightarrow (h,k)$
$P\left(\frac{1+a}{2},\frac{1-a}{2}\right)\Rightarrow (2h,2k)$

Equation of chord $\Rightarrow T=S_1$
$\Rightarrow (x-y)\lambda -\frac{2h(x+\lambda )}{2}-\frac{(2k)(y-\lambda )}{2}$
$=2{\lambda }^2-2h(\lambda )+2k\lambda$
Now, $\lambda (2h,2k)$ satisfies the chord
$\therefore (2h-2k)\lambda -h(x+\lambda )-k(y-\lambda )$
$\Rightarrow 2{\lambda }^2+4k\lambda -4h\lambda +h\lambda -k\lambda +hx+ky=0$
$\Rightarrow 2{\lambda }^2+\lambda (3k-3h)+ky+hx=0$
$\Rightarrow D>0$
$\Rightarrow 9(k-h{)}^2-8(ky+hx)>0$
$\Rightarrow 9(k-h{)}^2-8\left(2k^2+2h^2\right)>0$
$\Rightarrow -7k^2-7h^2-18kh>0$
$\Rightarrow 7k^2+7h^2+18kh<0$
$\Rightarrow 7{\left(\frac{1-a}{4}\right)}^2+7{\left(\frac{1+a}{4}\right)}^2+18\left(\frac{1-a^2}{16}\right)<0$
$\Rightarrow 7\left[\frac{2\left(1+a^2\right)}{16}\right]+\frac{18\left(1-a^2\right)}{16}<0,a^2=t$
$\Rightarrow \frac{7}{8}(1+t)+\frac{18(1-t)}{16}<0$
$\Rightarrow \frac{14+14t+18-18t}{16}<0$
$\Rightarrow 4t>32$
$t>8a^2>8$