Three equal circles $C_1,C_2$ and $C_3$ with center $O_1,O_2$ and $O_3$ respectively and radius $r$ have a common point $O$.Circle $C_1$ and $C_2$,$C_2$ and $C_3$, $C_3$ and $C_1$ meet again at point $A,B,C$ respectively.Prove that circumradius of the triangle $ABC$ is $r$.
It is not difficult to see that $OO_2BO_3$ is a rhombus.
This in turn makes $AO_2BX$ (where X is point of intersection of OA and the perpendicular bisector of AB) a rhombus.
Note that X is also the circum-center of $\triangle ABC$. The required circum-radius $= AX = O_2B = r$.
It is in fact linked to Johnson's circles (https://en.wikipedia.org/wiki/Johnson_circles)
Here is a proof using a constructive reasoning (found when I wanted to draw the figure with Geogebra).
Have a look at the following picture:
Consider that circles $(C_1)$ and $(C_2)$ have already been built with intersection points $A$ and $O$; how can $(C_3)$ be placed ? As $O$ must be a common point to the 3 circles, necessarily, the center $A_3$ must be chosen at distance $r$ from $O$. But this means that $A_3$ must belong to the (dotted) circle with center $O$ and radius $r$. Thus, the circumscribed circle to $\triangle A_1A_2A_3$ has necessarily radius $r$.