Although probably not very interesting for anyone, here we show how one can create the Typhon OS logo with just a compass and a rule:

Construction

The construction starts with a circle of radius 16¹

Now we will make the cutouts for the wing. This are achieved by another two circles, the top is made by a circle of radius 10 placed on the line $x = 0$ and that is a interior tangent to the outside circle. Therefore, the center is at $\left(0, 6\right)$. The bottom cutout is made in a similar fashion, but this time, the circle has radius 4 and it is placed in the line $y = x$. If you do the math, this makes it so the center is at $\left(-6\sqrt{2},-6\sqrt{2}\right)$.

Next, we make the dragon’s neck, this is done by (you guessed it) another circle. This one is of radius 5 and it is also placed in the line $y = x$ and it is placed so that it is tangent on the bottom to the neck’s circle’s cutout. That is, it’s center is at $\left(0,3\right)$

We now start with the head construction, don’t worry, there are no circles involved in it. Start from the intersection of the neck circle with the positive part of $x = 0$ (point $(0,6)$) and draw a horizontal segment of length 4. From there (point $(-4,6)$) draw a line at $\frac{\pi}{4}$ radians (45 degrees) angle. Draw the line $y = 2x$

Now, to make the body. We will draw a big circle of radius 14 and centered at the origin.

Cut the last circle with the fin line on the top right and with the cutout on the bottom left to obtain the basic shape of the logo.

Fins

We come to the most complicated part of the logo: the fins. The three of them are made in the exact same way but the construction is rotated for each of them.

Start with the line $y = 0$ and draw a circle of radius 4 with center on that line that is interior tangent to the original circle of radius 16. That makes the center be at $(12,0)$.

Now rotate the line around the center of the circle $\frac{\pi}{4}$ radians (45 degrees).

And draw a circle centered at the lowest intersection of this new line and the previous circle and that is tangent to the exterior circle of radius 16. If you do the math, the intersection point comes out to be $\left(2(6-\sqrt{2}, -2\sqrt{2}\right)$ and the radius $16-4\sqrt{10-3\sqrt{2}}$.

We can remove the auxiliary outside circle and the auxiliary lines.

And the fin is made by subtracting the smallest minus the bigger circle from the body.

Now we can repeat the same procedure with the line $y= -x$. In this case, the center of the smaller circle is at $(6\sqrt{2},-6\sqrt{2})$. The center of the bigger one is at $(-4+6\sqrt{2},-6\sqrt{2})$. Finally, by simmetry, the radius of the bigger circle is the same as for the first fin

Giving us

Finally we repeat the same steps for the line $x=0$. From that, we get the center of the smaller circle at $(0,-12)$. And the center of the bigger at $(-2\sqrt{2},-12+2\sqrt{2})$. And again, by simmetry, the radius is the same as the other fins.

Giving us finally