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If a point $\mathrm{P}(\alpha, \beta)$ on the line $\mathrm{y}=1$ is such that the two distinct chords drawn on $x^2+y^2-\alpha x-y=0$ from $P$ are bisected by the $x$-axis, then
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The correct answer is:
$\alpha^2>8$
Given $\mathrm{p}=(\alpha, \beta)$ and Let $\mathrm{C}$ is the mid point of chord PQ,
Let $\mathrm{C}=(\lambda, 0) \Rightarrow \beta=(-\alpha+2 \mathrm{~h},-\beta)$
Since $\beta$ lies on the circle.
$\begin{aligned} & \mathrm{x}^2+\mathrm{y}^2=\alpha \mathrm{x}+\mathrm{y} \\ & \Rightarrow(-\alpha+2 \mathrm{~h})^2+(-\beta)^2=(-\alpha+2 \mathrm{~h})+(\beta)\end{aligned}$
$\begin{aligned} & \text { or } \Rightarrow(-\alpha+2 \mathrm{~h})^2+(-1)^2=\alpha(-\alpha+2 \mathrm{~h})-1(\because \beta=1) \\ & \Rightarrow 2 \mathrm{~h}^2-3 \alpha \mathrm{h}+\alpha^2+1=0\end{aligned}$
Since there are two distinct chords bisected by $\mathrm{x}$-axis then for $h$ to be real
$\begin{aligned} & (-3 \alpha)^2-4.2\left(\alpha^2+1\right)>0 \\ & \Rightarrow \alpha^2>8\end{aligned}$
Let $\mathrm{C}=(\lambda, 0) \Rightarrow \beta=(-\alpha+2 \mathrm{~h},-\beta)$
Since $\beta$ lies on the circle.
$\begin{aligned} & \mathrm{x}^2+\mathrm{y}^2=\alpha \mathrm{x}+\mathrm{y} \\ & \Rightarrow(-\alpha+2 \mathrm{~h})^2+(-\beta)^2=(-\alpha+2 \mathrm{~h})+(\beta)\end{aligned}$
$\begin{aligned} & \text { or } \Rightarrow(-\alpha+2 \mathrm{~h})^2+(-1)^2=\alpha(-\alpha+2 \mathrm{~h})-1(\because \beta=1) \\ & \Rightarrow 2 \mathrm{~h}^2-3 \alpha \mathrm{h}+\alpha^2+1=0\end{aligned}$
Since there are two distinct chords bisected by $\mathrm{x}$-axis then for $h$ to be real
$\begin{aligned} & (-3 \alpha)^2-4.2\left(\alpha^2+1\right)>0 \\ & \Rightarrow \alpha^2>8\end{aligned}$
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