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As visually discussed here, a set of equations has recently been popping up as graffiti in Belgium. The equations define five functions of one variable, namely:

$$\begin{array}{rcl}f\left(x\right)& =& 2+\sqrt{-{(x-2)}^{2}+1}\\ g\left(x\right)& =& 2-\sqrt{-{(x-2)}^{2}+1}\\ h\left(x\right)& =& 3x-3\\ i\left(x\right)& =& -3x+9\\ j\left(x\right)& =& \mathrm{0,2}x+\mathrm{1,7}\end{array}$$

Plotting the five functions with $x\in [1,3]$ (the domain of existence of $f$ and $g$) gives the well-known anarchist logo

As mathematicians, we can take this a step further and define an
**Anarchist curve**, by finding the implicit form of the plot of each
of the function, and then bringing them together.

In this case, $f,g$ together define the circle, with equation

$${(x-2)}^{2}+{(y-2)}^{2}=1$$

or rather (fully implicit):

$${(x-2)}^{2}+{(y-2)}^{2}-1=0.$$

The three functions $h,i,j$ describe the ‘A’ shape. We first rewrite $j$ in a nice form as

$$j\left(x\right)=x/5+17/10,$$

and then write the implicit equation for each of them, multiplying the one for $j$ by $10$ to get rid of the fractions:

$$\begin{array}{rcc}h& :& y-3x+3=0\\ i& :& y+3x-9=0\\ j& :& 10y-2x-17=0\end{array}$$

We can now multiply all the left hands together, obtaining:

$$(y-3x+3)(y+3x-9)(10y-2x-17)=0$$

which is the equation for the ‘A’.

If we then multiply this for the left-hand side of the implicit
equation for the circle, we have the **Anarchist curve**

$$({(x-2)}^{2}+{(y-2)}^{2}-1)(y-3x+3)(y+3x-9)(10y-2x-17)=0$$

(which, for the record, is currently missing from the list of known curves in the Wolfram Alpha database.)

(There's a few more we could draw similarly, but that's for another time.)