How to show that the graph $y^2=\sin{\frac{\pi}{2}x}$ does not consist of circles?

The question in the textbook gives the graph of $y^2=\sin{\frac{\pi}{2}x}$ which looks like this.

It then asks to prove that although the graph looks like it is made from circles, it is not. I'm not sure how to prove this exactly. I tried to show this by finding the derivative which is $\frac{dy}{dx}=\frac{\pi\cos{(\pi/2)x}}{4y}$. Then I thought since the numerator isn't a linear function it wouldn't be a circle but I'm not sure if that conclusion makes any sense. Could someone clarify how could I disprove that the graph is made of circles?

Here's a more geometric approach. The pieces are identical since $\sin\left(\frac{\pi}{2}x\right)$ is periodic so it's sufficient to show any one of them is not a circle. Consider $x\in[0,2]$ and suppose to the contrary that the graph were a circle with center $O$. Points $A=(0,0)$ and $B=(2,0)$ are on the graph so $O$ must lie on the perpendicular bisector of segment $AB$, which is the line $x=1$. Similarly, $C=(1,1)$ and $D=(1,-1)$ are on the graph so O must be on the line $y=0$, the perpendicular bisector of segment $CD$. Thus, the lines $x=1$ and $y=0$ intersect at $O$, i.e. $O=(1,0)$, and the radius is equal to $OA=1$. Point $E=\left(\frac 13,\frac{1}{\sqrt 2}\right)$ is on the graph as well. From the distance formula, we have $$OE=\sqrt{\left(1-\frac 13\right)^2+\left(\frac{1}{\sqrt 2}\right)^2 }=\sqrt{\frac{17}{18}}\ne 1,$$ which is a contradiction.