Q:

a problem on a given set and its connected components

For $v ∈ \mathbb{R}^2$ and $r > 0$  let $D(v, r)$  denote the closed disc with centre at $v$
and radius $r$. Let $v = (5, 0) ∈\mathbb{R} ^2$. For $α > 0$ let   $X_α$ be the subset
$X_α = D (−v, 3) ∪ D(v, 3) ∪$ {$(x, αx) : x ∈\mathbb{R}$}. 
Determine the condition on α for $X_α$ to be connected; when it is not con-nected how many connected components does $X_α$ have? 

clearly if $X_α$ is disconnected then it must have three components. 
the main problem is to finding the value of $\alpha$. $(x, αx)$ is a straight line passing through the origin.we need to find such $\alpha$ that the straight line intersects both the disk.  but I can not solve $\alpha$.can anybody help me.  

A:

The equation for boundary of $D(v,3)$ for instance is $$(x-5)^2 + y^2 = 3^2$$ Now you have to find $\alpha$ such that after subsituting $y$ with $\alpha x$ this equation will have at least one solution (which means that line $y=\alpha x$ and this circle intersects)
After subsitution we have $$(x-5)^2 + \alpha^2x^2 = 9$$ which expands to $$x^2(\alpha^2 + 1) - 10x + 16 = 0$$ It has a solution iff it's discriminant is non negatve.  We calculate $$\Delta =(-10)^2 - 4(\alpha^2 + 1)\cdot 16 = 100 - 64 - 64\alpha^2 = 36 - 64\alpha^2$$ and then we need to solve $\Delta \ge 0$ which yields $$\alpha^2 \le\frac{36}{64} = \frac{9}{16} = (\frac{3}{4})^2$$
And we get a solution $|\alpha| \le \frac{3}{4}$. Looking at the second disc will give you the same solution, so this line intersects with none or both of discs.  And finally we have that for $\alpha \in [-\frac{3}{4}, \frac{3}{4}]$ there is one connected component, and for $\alpha \notin [-\frac{3}{4}, \frac{3}{4}]$ there are three connected components.