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

Consider the following shape which is produced by dividing the line between $0$ and $1$ on $x$ and $y$ axes into $n=16$ parts.

Question 1: What is the curve $f$ when $n\rightarrow \infty$?

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Update: According to the answers this curve is not a part of a circle but with a very similar properties and behavior. In the other words this fact shows that how one can produce a "pseudo-cycle" with equation $x^{\frac{1}{2}}+y^{\frac{1}{2}}=1$ from some simple geometric objects (lines) by a limit construction.

Question 2: Is there a similar "limit construction by lines" like above drawing for producing a circle?

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 $\pi$!

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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. $\qquad\square$

Hint: find the general equation for the lines.

Then, find a formula for "the point $(x,y)$ lies above and to the right of a given line".

To find the general equation for the lines, we can use the description that we've divided the axes into equal pieces. If there are $n$ equal pieces, then there are $n-1$ points in those intervals $[0,1]$, located at distances $m/n$ from the origin, where $m$ is a positive integer less than $n$.

The line attached to $(m/n, 0)$ has $(0, (n-m)/n) = (0, 1 - m/n)$ as its other end point. (There are lots of ways to see this: I'll let you choose your own!)

To save writing, I'll let $a = m/n$, so the line segments connect $(a,0)$ to $(0, 1-a)$.

Now, we can any method for obtaining the equation of a line through $(a,0)$ and $(0, 1-a)$. We could use the two-point form. We could use slope intercept form. I'll demonstrate a method that doesn't rely on actually remembering something: the equation of a line always looks like

$$ c x + d y = e $$

for some $c$, $d$, and $e$. Since those two points have to lie on it, we have a system of equations:

$$ ca + d0 = e \qquad \qquad c0 + d(1-a) = e $$

or more simply

$$ ac = e \qquad \qquad d(1-a) = e $$

and we just need any solution that is not all zeroes. One such solution is

$$ c = \frac{1}{a} \qquad \qquad d = \frac{1}{1-a} \qquad \qquad e = 1 $$

Aside: we could have taken a different starting point: e.g. we could have said the equation of a non-vertical line always looks like $y = cx + d$, and taken the same approach to figure out what $c$ and $d$ are.

So, the equations for the line all look like

$$ \frac{x}{a} + \frac{y}{1-a} = 1 $$

One side of the line is given by the inequality

$$ \frac{x}{a} + \frac{y}{1-a} > 1 $$

and the other by

$$ \frac{x}{a} + \frac{y}{1-a} < 1 $$

We want the inequality that describes points being above/to the right of the line. While you could try to "reason it out", an easier approach is just to observe that we want the side containing $(1,1)$. An even easier approach is to observe we don't want the side containing $(0,0)$. If we plug in $(0,0)$ into the left hand side, we get zero. So we don't want the second one.

Thus, every point on and above your beauty curve satisfies the inequality

$$ \frac{x}{a} + \frac{y}{1-a} \geq 1 $$

for every rational value of $a \in (0,1)$. It really ought to satisfy the inequality for every $a \in (0,1)$ too.

If your curve behaves as nicely as it looks, this inequality should be an actual equality for exactly one value of $a$.

So which points $(x,y)$ satisfy this inequality for all values of $a$? Well, let's solve for which $a$'s it does satisfy!

First combine the fractions:

$$ \frac{x(1-a) + ya}{a(1-a)} \geq 1 $$

Now multiply through by the denominator. Don't forget to check the sign! It's positive, so that means the inequality stays in the same direction. So, we get

$$ x(1-a) + ya \geq a(1-a) $$

collecting the $a$'s together:

$$ a^2 + a(y-x-1) + x \geq 0 $$

(and we also have $0 < a < 1$). This is a quadratic function of $a$: for each particular value of $x,y$, it's graph is a parabola.

Since we want the points actually on the beauty curve, we desire that the actual equation

$$ a^2 + a(y-x-1) + x = 0 $$

have exactly one solution for $a$. Recall that the solutions are, by the quadratic formula,

$$ a = \frac{(1+x-y) \pm \sqrt{(1+x-y)^2 - 4x}}{2} $$

If there is only one solution, then that square root must actually be zero. That is, we need

$$ (1+x-y)^2 - 4x = 0 $$

or equivalently

$$ x^2 - 2xy + y^2 - 2x - 2y + 1 = 0 $$