When does the union of two Disks produce a convex set?

Question

I have just graced the subject of optimization, am doing the MIT OCW 18.065 course, and one of the hw is on convexity, Could someone help? I think the solution is if you overlay one disk over the other, that would be a convex set? ie one is a subset of the other

Answer 1

Hint: what does the union of two disks look like near a point (if it exists) where their boundaries intersect? What if the boundaries don't intersect?

EDIT: For squares, this is convex:

and so is this:

And of course if one square is contained in the other. But I don't believe there are any other convex cases.

Answer 2

The union of two disks is a convex set if and only if one is a subset of the other.

(The union of two circles is never convex.)

Answer 3

Hint A: Consider tangent lines. What can these tell you about convexity of disks?

Hint B: Is there a way you can adopt the idea behind your answer to Hint A to give you some insight into squares?

Answer 4

We solve it in the more general, $n-$dimensional case. Assume the sphere's are $\Vert y-x_1\Vert \le r_1$ and $\Vert z-x_2\Vert \le r_2$. For the new point $\lambda y+(1-\lambda)z$ living on the chord that connects $y,z$, one of the following conditions must be true (because we are taking the union) (and also $\lambda \in [0,1]$)

$$\begin{cases}\Vert \lambda y+(1-\lambda)z-x_1\Vert \le r_1 \\ \Vert \lambda y+(1-\lambda)z-x_2\Vert \le r_2\end{cases} $$

consider for example the the first equation, $\Vert \lambda y+(1-\lambda)z-x_1\Vert \le r_1 \Rightarrow \Vert \lambda (y-x_1)+(1-\lambda)(z-x_1)\Vert \le \lambda\Vert y-x_1\Vert+(1-\lambda)\Vert z-x_1\Vert \le \lambda r_1 +(1-\lambda)\Vert z-x_1\Vert \Rightarrow (1-\lambda)\Vert z-x_1\Vert \le r_1-\lambda r_1 \Rightarrow \Vert z-x_1\Vert \le r_1$

This means that $z$ must satisfy both $\Vert z-x_2\Vert \le r_2$ and $\Vert z-x_1\Vert \le r_1$. same goes for $y$ too, if you do the similar chain of reasoning, starting from the second equation. This means each point of one of the spheres must be in the other one as well. Therefore the union of two spheres is convex iff one sphere is a subset of the other.