Mathematics - Circle Question with Solution | TestHub

Given,

The line $y=-x$ bisect the chord $PQ$

If $\left(\lambda ,-\lambda \right)$ is the mid-point of the chord $PQ$ then,

$Q(2\lambda -\alpha ,-2\lambda -\beta )$

Given circle $2x^2+2y^2-(1+a)x-(1-a)y=0$

Or, $x^2+y^2-\left(\frac{1+a}{2}\right)x-\left(\frac{1-a}{2}\right)y=0$

Let $\frac{1+a}{2}=\alpha ,\frac{1-a}{2}=\beta$

The circle equation will be,

$x^2+y^2-\alpha x-\beta y=0$

Point $Q$ will lie on the circle,

$(2\lambda -\alpha {)}^2+(-2\lambda -\beta {)}^2-\alpha (2\lambda -\alpha )-\beta (-2\lambda -\beta )=0$

$\Rightarrow 4{\lambda }^2+3\lambda (\beta -\alpha )+\left({\alpha }^2+{\beta }^2\right)=0$

So, quadratic in $\lambda$ should have $D>0$

$\Rightarrow 9(\beta -\alpha {)}^2-4(4)\left({\alpha }^2+{\beta }^2\right)>0$

Put the value of $\alpha$ and $\beta$ we get,

$\Rightarrow 9a^2-16\left(\frac{1+a^2}{2}\right)>0$

$\Rightarrow a^2>8$