# Obtaining condition for existence of two Chords through a single point with their Mid-points on the $x$-Axis

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.