Question

Find the intervals of values of ${}^{\prime }{a}^{\prime }$ 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 $(2x^2+2y^2-(1+\sqrt{2}a)x-(1+\sqrt{2}a\left)y\right)=0$

• A. $a\in (-\infty ,-2)\cup (2,\infty )$
• B. $a\in (-2,2)$
• C. $a\in (-\infty ,-4)\cup (4,\infty )$
• D. $a\in [-2,2]$

Answer 1

The correct option is A $a\in (-\infty ,-2)\cup (2,\infty )$

$2x^2+2y^2-(1+\sqrt{2}a)x-(1-\sqrt{2}a)y=0$
$\Rightarrow$ $x^2+y^2-\left(\frac{1+\sqrt{2}a}{2}\right)x-\left(\frac{1-\sqrt{2}a}{2}\right)y=0$
Since $y+x=0$ bisects two chords of this circle, mid-points of the chords must be of the form $(\alpha ,-\alpha )$
Equation of the chord having $(\alpha ,-\alpha )$ as mid-points is $T=S_1$
$\Rightarrow$ $x\alpha +y(-\alpha )-\left(\frac{1+\sqrt{2}a}{4}\right)(x+\alpha )-\left(\frac{1-\sqrt{2}a}{4}\right)(y-\alpha )$
$={\alpha }^2+{-\alpha }^2-\left(\frac{1+\sqrt{2}a}{2}\right)\alpha -\left(\frac{1+\sqrt{2}a}{2}\right)(-\alpha )$
$\Rightarrow$ $4x\alpha -4y\alpha -(1-\sqrt{2}a)x-(1+\sqrt{2}a)\alpha -(1-\sqrt{2}a)y+(1-\sqrt{2}a)\alpha$
$=4{\alpha }^2+4{\alpha }^2-(1+\sqrt{2}a)y).2.\alpha +(1-\sqrt{2}a)y$
$=8{\alpha }^2-(1+\sqrt{2}a)\alpha +(1-\sqrt{2}a)\alpha$
But this chord will pass through the point
$(\frac{1+\sqrt{2}a}{2},\frac{1-\sqrt{2}a}{2})$
$\therefore$ $4\alpha \left(\frac{1+\sqrt{2}a}{2}\right)-4\alpha \left(\frac{1-\sqrt{2}a}{2}\right)-\frac{(1+\sqrt{2}a)(1+\sqrt{2}a)}{2}-\frac{(1-\sqrt{2}a)(1-\sqrt{2}a)}{2}$
$=8{\alpha }^2-2\sqrt{2}a\alpha$
$\Rightarrow$ $2\alpha \left[\right(1+\sqrt{2}a-1+\sqrt{2}a\left)\right]=8{\alpha }^2-2\sqrt{2}a\alpha$
$\Rightarrow$ $4\sqrt{2}a\alpha -\frac{1}{2}[2+2{\left(\sqrt{2}a\right)}^2]=8{\alpha }^2-2\sqrt{2}a\alpha$
[$\because$ ${(a+b)}^2+{(a-b)}^2=2a^2+2b^2$]
$\Rightarrow$ $8{\alpha }^2-6\sqrt{2}a\alpha +1+2a^2=0$
But this quadratic equation will have two distinct roots, if ${\left(6\sqrt{2}\right)}^2-4\left(8\right)(1+2a^2)>0$
$\Rightarrow$ $72a^2-32(1+{2a}^2)>0$
$\Rightarrow$ $72a^2-32-64a^2>0$ $\Rightarrow$ $8a^2-32>0$
$\Rightarrow$ $a^2>4$ $\Rightarrow$ $a<-2\cup a>2)$
Therefore, $a\in (-\infty ,-2)\cup (2,\infty )$