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(C) Given the point $P(\alpha, 1)$ lies on the line $y=1$. Let the chord be $PQ$ and its midpoint be $M(h, 0)$ on the $x$-axis.The equation of the cho

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$\alpha^2 > 8$

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(C) Given the point $P(\alpha, 1)$ lies on the line $y=1$. Let the chord be $PQ$ and its midpoint be $M(h, 0)$ on the $x$-axis.The equation of the chord of the circle $x^2+y^2-\alpha x-y=0$ with midpoint $(h, k)$ is $T=S_1$.Here,$T = xh + yk - \frac{\alpha}{2}(x+h) - \frac{1}{2}(y+k)$ and $S_1 = h^2+k^2-\alpha h-k$.Since the midpoint is $(h, 0)$,we have $T = xh - \frac{\alpha}{2}(x+h) - \frac{1}{2}y$ and $S_1 = h^2-\alpha h$.So,$xh - \frac{\alpha}{2}x - \frac{\alpha}{2}h - \frac{1}{2}y = h^2-\alpha h$.Since this chord passes through $P(\alpha, 1)$,we substitute $x=\alpha$ and $y=1$:$\alpha h - \frac{\alpha^2}{2} - \frac{\alpha h}{2} - \frac{1}{2} = h^2 - \alpha h$.Multiplying by $2$: $2\alpha h - \alpha^2 - \alpha h - 1 = 2h^2 - 2\alpha h$.Rearranging terms: $2h^2 - 3\alpha h + \alpha^2 + 1 = 0$.For two distinct chords to exist,the quadratic equation in $h$ must have two distinct real roots.Thus,the discriminant $D > 0$:$(-3\alpha)^2 - 4(2)(\alpha^2+1) > 0$.$9\alpha^2 - 8\alpha^2 - 8 > 0$.$\alpha^2 - 8 > 0 \Rightarrow \alpha^2 > 8$.

If a point $P(\alpha, \beta)$ on the line $y=1$ 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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