Question

The range of value 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 $2x^2+2y^2-(1+\sqrt{2}a)x-(1-\sqrt{2}a)y=0$ is

• A. $(-\infty ,-2)\cup (2,\infty )$
• B. $(-2,2)$
• C. $(2,\infty )$
• D. None of these

Answer 1

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

Equation of the given circle can be written as $x^2+y^2-\left(\frac{1+\sqrt{2}a}{2}\right)x-\left(\frac{1-\sqrt{2}a}{2}\right)y=0.$

Since $y+x=0$ bisects two chord of this circle, mid-points of the chords must be the form $(\alpha ,-\alpha ).$

Equation of the chord having $(\alpha ,-\alpha )$ as mid-point is $T=S_1$ i.e., $\alpha x+(-\alpha )y-\left(\frac{1+\sqrt{2}a}{4}\right)(x+\alpha )-\left(\frac{1-\sqrt{2}a}{4}\right)(y+\alpha )$

$={\alpha }^2+{\alpha }^2-\left(\frac{1+\sqrt{2}a}{2}\right)\alpha -\left(\frac{1-\sqrt{2}a}{2}\right)(-\alpha )$

$\Rightarrow 4\alpha x-4\alpha y-(1+\sqrt{2}a)x-(1-\sqrt{2}a)y=8{\alpha }^2-(1+\sqrt{2}a)\alpha +(1-\sqrt{2}a)\alpha .$

This chord will pass through the point $(\frac{1+\sqrt{2}a}{2},\frac{1-\sqrt{2}a}{2})$ if

$4\alpha \left(\frac{1+\sqrt{2}a}{2}\right)-4\alpha \left(\frac{1-\sqrt{2}a}{2}\right)-\frac{(1+\sqrt{2}a)(1+\sqrt{2}a)}{2}-\frac{(1-\sqrt{2}a)(1-\sqrt{2}a)}{2}$

$=8{\alpha }^2-2\sqrt{2}a\alpha$

$\Rightarrow 2\alpha [1+\sqrt{2}a-1+\sqrt{2}a]-\frac{1}{2}[{(1+\sqrt{2}a)}^2+{(1-\sqrt{2}a)}^2]=8{\alpha }^2-2\sqrt{2}a\alpha$

$\Rightarrow 4\sqrt{2}a\alpha -\frac{1}{2}[2{\left(1\right)}^2+2{\left(\sqrt{2}a\right)}^2]=8{\alpha }^2-2\sqrt{2}a\alpha$

$[\because {(a+b)}^2+{(a-b)}^2=2a^2+2b^2]$

$\Rightarrow 8{\alpha }^2-6\sqrt{2}a\alpha +1+2a^2=0.$

This quadratic equation will have two distinct roots if

${\left(6\sqrt{2}a\right)}^2-4\left(8\right)(1+2a^2)>0.$

$\Rightarrow 72a^2-32(1+2a^2)>0\Rightarrow 9a^2-4-8a^2>0$

$\Rightarrow a^2-4>0\Rightarrow (a-2)(a+2)>0$

$\Rightarrow a\in (-\infty ,-2)\cup (2,\infty ).$