Producing circles with $r^2=x^2+y^2$ $r^2=x^2+y^2\ +\pi $ produces graph with two circles :
$r = 2$

This graph was produced using Desmos, is this correct or a quirk of the graph software ?
Ive noticed $\forall n \epsilon N [r^2=x^2+y^2+n\pi]$ appears to produce same circle, so $r^2=x^2+y^2\ +3\pi $ produces same circle as $r^2=x^2+y^2\ +4\pi $  :

Why is this the case, should each circle not be differing sizes ?
 A: The equation $2^2=x^2+y^2+\pi$ is equivalent to $x^2+y^2=4-\pi=0.8584\ldots$, which produces a circle of radius $\sqrt{4-\pi}=\sqrt{0.8584\ldots} = 0.9265\ldots$, which is the red circle in the top figure.
The equation $2^2=x^2+y^2+3\pi$ is equivalent to $x^2+y^2=4-3\pi=-5.4247\ldots$. Since that's a negative number, there are no solutions to the equation for real numbers $x$ and $y$. So there is no red circle in the bottom figure.
I don't know what the orange circle means in each figure. Try consulting the software's manual?
A: Your first picture is correct if you are graphing
\begin{align*}
2^2 &= x^2 + y^2\\
- \pi + 2^2 &= x^2 + y^2
\end{align*}
for red and yellow respectively. If we have the equation
$$
r^2 = x^2 + y^2 + c
$$
where $c$ is a constant, that is the same as
$$
r^2 - c = x^2 + y^2
$$
and so our new radius is $r' = \sqrt{r^2 - c}$, if $c > r^2$, the circle has no points as squares are positive. As $3\pi$ and $4\pi$ are larger than $2^2$, there are no solutions for
\begin{align*}
-3\pi + 2^2 = x^2 + y^2\\
-4\pi + 2^2 = x^2 + y^2.
\end{align*}
Desmos will therefore not draw anything.
A: First note that $r^2=x^2+y^2+n\pi$ produces circles of radius equal to $\sqrt{r^2-n\pi}$. So you need $r^2>n\pi$ to have one. In the case $r=2$, the only possible value is $n=1$. Desmos cannot draw any other case because there is no other case!
Instead, if you use $r^2=x^2+y^2-n\pi$, then Desmos draws the corresponding circle of radius $\sqrt{r^2+n\pi}$ for any $r,n\in\mathbb{R}$. I tried it myself and it seems to we working properly. My guess is that you should revise the commands that you are using...
