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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.

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So, given your analysis, what remains is merely a problem of coordinate Euclidean geometry, right? –  Hurkyl Jan 4 '13 at 9:02

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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.

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