What is this curve formed by latticing lines from the $x$ and $y$ axes?

If we attach four curves like $f$ to each other in the following form a "pseudo-circle" shape appears. Note to its difference with a real circle. Its formula is $x^{\frac{1}{2}}+y^{\frac{1}{2}}=1$ a dual form of the circle equation $x^{2}+y^{2}=1$. You can find this equation simply by the geometric analysis of each line.

A very interesting point about this curve is that there is a kind of $\pi$ for it which doesn't change by radius! Here we have $\pi'=\frac{10}{3}=3.3333...$ which is very near to the $\pi$ of circle ($=3.1415...$) but $\pi'$ is a rational number not a non-algebraic real number like $\pi$!

Each line connects $(a,0)$ with $(0,1-a)$, so has equation $y=(1-a)-\frac{1-a}ax$

At a given $x$, we can find the $a$ that results in the highest $y$ by differentiating: $\frac {dy}{da}=-1-x\frac{-a-(1-a)}{a^2}=-1+\frac x{a^2}$ which is zero when $a=\sqrt x$

Plugging this in, we get the curve $y=1-2\sqrt x +x$, plotted here by Alpha

This can be expressed as $\sqrt x + \sqrt y = 1,$ which is nicely symmetric.

It is a parabola rotated through an eighth of a circle, satisfying $$(x+y) = \dfrac{(x-y)^2}{2} + \frac12$$ or $$x^2+y^2 -2xy-2x-2y+1=0$$ which has the solution in this part of the curve of $$y=(1-\sqrt{x})^2 \,\text{ i.e. }\, \sqrt{x}+\sqrt{y}=1 $$ though with other solutions in other parts of the same parabola extended. Its derivative is therefore $$\frac{dy}{dx}=-\frac{1-\sqrt{x}}{\sqrt{x}}$$ and so the tangent at $(x,(1-\sqrt{x})^2)$ crosses the $y$-axis at $(0, 1 - \sqrt{x}) $ and the $x$-axis at $(\sqrt{x},0)$: since $1 - \sqrt{x}+\sqrt{x}=1$, this satisfies the original construction. $\qquad\square$