Given a circle $x^2+y^2=\alpha x + \beta y$, obtain the condition on $\alpha$ and $\beta$, if two distinct chords can be drawn through the point $(\alpha, \beta)$ such that their mid-points lie on the $x$-axis.
For starters, it is evident that $A(\alpha, \beta)$ will lie on the circle. I have introduced a parameter $\theta$ to represent all lines passing through $A$. Also, we need to make sure that every such line intersects the $x$-axis inside the circle, as it is the mid-point that must be the intersection point, and therefore we must have $\tan^{-1} \frac{\beta}{\alpha} \lt \theta \lt \frac{\pi}{2}$. The equation of any such line is $$x=y\cot\theta +\alpha -\beta \cot\theta$$ Putting this value of $x$ in the circle’s equation, I can get the $y$-coordinate of the other end of the chord: $$y=(\beta \cot \theta -\alpha)\sin\theta\cos\theta $$ And so the ordinate of the midpoint can be found and equated to zero. $$\frac{(\beta\cot\theta-\alpha)\sin\theta\cos\theta + \beta}{2} =0 $$ This can be transformed into a quadratic in $\cos^2\theta$. $$(\alpha^2+\beta^2)\cos^4\theta+(2\beta^2-\alpha^2)\cos^2\theta+\beta^2=0 $$ According to the question, it is required that this equation have two distinct roots, and so $D\gt0$, i.e. $$(2\beta^2-\alpha^2)^2 -4\beta^2(\alpha^2+\beta^2) \gt 0 \\ \implies |\alpha| \gt 2\sqrt 2 |\beta|$$
I would like to ask if my solution is correct, and is there a faster and more efficient way? How would you tackle this problem?
If the midpoint of a chord with one endpoint at $(\alpha,\beta)$ lies on the $x$-axis, then the $y$-coordinate of the other endpoint must be $-\beta$. We therefore seek $\alpha$ and $\beta$ such that $x^2+\beta^2=\alpha x-\beta^2$ has two distinct real solutions. The discriminant of this quadratic equation is $\alpha^2-8\beta^2$, from which we obtain the condition $\alpha^2\gt8\beta^2$. This agrees with your result.