There is a disk of radius $r(x)=2-\frac{2}{5}|x|$ at every point on $[-5, 5]$. Find equation of a union of all disks Modeling shows that the figure look like this:

But I don't know how to get an equation for it analytically
Looks like for some $0<x_0<5$ for $(x_0, 5]$ and $[-5, x_0)$ correspoing disks touch final shape at only one point each. It would help to find at least this $x_0$.
 A: Based on the diagram and the comment from @ajotatxe, I will assume that you meant to write $r(x) = 2 - \frac{2}{5}|x|$. Fix a point $a$ on the interval $[0,5]$ and consider the disk
$$(x-a)^2 + y^2 = r(a)^2 =  \left(2 - \frac{2}{5}a\right)^2 = 4 - \frac{8}{5}a + \frac{4}{25}a^2$$
centered at $a$. There is a unique coordinate $x_a$ so that the tangents to this circle at $(x_a,y_a)$ passes through the point $(5,0)$, where $y_a = \sqrt{r(a)^2-(x_a-a)^2}$. To find it, we calculate
$$2(x-a) + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x-a}{y}$$
and so $x_a$ satisfies
$$0 - y_a = -\frac{x_a - a}{y_a}(5 - x_a) \implies (x_a-a)(5-x_a) =y_a^2 = r(a)^2 - (x_a-a)^2  $$
$$\implies -x_a^2 - 5a + (a+5)x_a = r(a)^2 - x_a^2 + 2ax_a - a^2 $$
$$\implies (5-a)x_a = r(a)^2 + 5a - a^2 = 4 + \frac{17}{5}a - \frac{21}{25}a^2 = \frac{-1}{25}(a-5)(21a+20) $$
and so $x_a = \frac{21}{25}a + \frac{4}{5}$. Thus $5-x_a = \frac{21}{25}(5 - a)$ and $x_a -a =\frac{4}{25}(5 - a)$, so $y_a = \pm\frac{2\sqrt{21}}{25}(5-a)$. This means the equation of the desired tangent line is $y - y_a = \mp\frac{2}{\sqrt{21}}(x-x_a)$. Notice that the slope of this line, $\frac{\mp 2}{\sqrt{21}}$, does not depend on $a$. Since by construction the tangent line always includes the point $(5,0)$, we see that they are all in fact the exact same line! Therefore, as the disks move from being centered at $(5,0)$ to $(0,0)$, they slide tangentially along the lines $y = \frac{\mp 2}{\sqrt{21}}(x-5)$.
Now since the biggest circle is at $a = 0$, we substitute to find that the last points in the shaded region along these common tangents will be at $(x_0,y_0) = \left(\frac{4}{5},\pm \frac{2\sqrt{21}}{5}\right)$, and on the interval $-\frac{4}{5} \leq x \leq \frac{4}{5}$ the upper and lower boundaries are the same as that of the circle $x^2 + y^2 = 4$. So to summarize,the region is:
$$(x,y) \text{ so that } \begin{cases} |y| \leq \sqrt{4 - x^2} & \text{ if } |x| \leq \frac{4}{5} \\  |y| \leq \frac{2}{\sqrt{21}}(5-x) & \text{ if } \frac{4}{5} \leq  |x| \leq 5 \end{cases}  $$

I also graphed this region in Desmos with a slider if you want to play around with it.
