What is the name of this curve?

The particular curve suggested, tangent to all of the segments from $(0,a)$ to $(9-a,0)$ as $a$ runs from $0$ to $9$, is this one: $$ y = x-6\sqrt{x}+9 $$

This is a portion of a parabola, with axis on the $x=y$ line and vertex at $(9/4,9/4)$.

To see this more clearly, add also such lines with $a>9$ and $a<0$ ...

For this particular example, the red curve is "stitched" by a family of straight lines of $$\begin{align*} \frac x{9-k} + \frac yk =& 1\\ y =& k - \frac{kx}{9-k} \end{align*}$$ where $0<k<9$.

When there are two such straight lines with parameters $h$ and $k$ respectively, $h\ne k$, $$\begin{align*} y =& h - \frac{hx}{9-h}\\ y =& k - \frac{kx}{9-k}\\ \end{align*}$$

Their intersection can be calculated as $$\begin{align*} h - \frac{hx}{9-h} =& k - \frac{kx}{9-k}\\ \frac{kx}{9-k} - \frac{hx}{9-h} =& k - h\\ x\cdot\frac{9k-hk-9h+kh}{(9-k)(9-h)} =& k - h\\ x =& \frac{(9-k)(9-h)}9\\ \end{align*}$$

When we take $h\to k$, the $x$-coordinates becomes $$x = \frac{(9-k)^2}9$$

and the $y$-coordinates is $$y = k - \frac{kx}{9-k} = k - \frac{k(9-k)}{9} = \frac{k^2}9$$

We can get an implicit curve for our range $$\sqrt x + \sqrt y = 3$$