JEE Main 2023 — Mathematics Coordinate Geometry

Given,

The line $y=-x$y=−x bisect the chord $PQ$PQ

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

$$Q(2\lambda -\alpha ,-2\lambda -\beta )$$Q(2λ−α,−2λ−β)

Given circle $2{x}^{2}+2{y}^{2}-(1+a)x-(1-a)y=0$2x2+2y2−(1+a)x−(1−a)y=0

Or, ${x}^{2}+{y}^{2}-(\frac{1+a}{2})x-(\frac{1-a}{2})y=0$x2+y2−(21+a​)x−(21−a​)y=0

Let $\frac{1+a}{2}=\alpha ,\frac{1-a}{2}=\beta$21+a​=α,21−a​=β

The circle equation will be,

$${x}^{2}+{y}^{2}-\alpha x-\beta y=0$$x2+y2−αx−βy=0

Point $Q$Q will lie on the circle,

$$(2\lambda -\alpha {)}^{2}+(-2\lambda -\beta {)}^{2}-\alpha (2\lambda -\alpha )-\beta (-2\lambda -\beta )=0$$(2λ−α)2+(−2λ−β)2−α(2λ−α)−β(−2λ−β)=0

$$\Rightarrow 4{\lambda }^{2}+3\lambda (\beta -\alpha )+({\alpha }^{2}+{\beta }^{2})=0$$⇒4λ2+3λ(β−α)+(α2+β2)=0

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

$$\Rightarrow 9(\beta -\alpha {)}^{2}-4(4)({\alpha }^{2}+{\beta }^{2})>0$$⇒9(β−α)2−4(4)(α2+β2)>0

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

$$\Rightarrow 9{a}^{2}-16(\frac{1+{a}^{2}}{2})>0$$⇒9a2−16(21+a2​)>0

$$\Rightarrow {a}^{2}>8$$⇒a2>8