Let $M$ be the midpoint of $B_1C_1$, $I$ the incenter and $O$ the circumcenter. It is well know that $B_1I=B_1A$ and $C_1A=C_1I$ so $B_1C_1$ is the perpedicular bisector of $AI$ so we get $IM=AM=A_2M$. And also $B_1C_1//IA_2$ as perpedicular to $AI$ now we get $OI=OA_2$ similar for $B_2,C_2$ and we done

Let $O$ be the circumcenter of triangle $ABC$. Then, $\angle OA_1B_2=\angle OA_1B-\angle B_2A_1B=90^\circ-\dfrac{\angle A}{2}-(180^\circ-\angle C_1BA_1)=90^\circ-\dfrac{\angle A}{2}-\angle C_1AA_1=$ $=90^\circ-\dfrac{\angle A}{2}-(\dfrac{\angle A+\angle C}{2})=\dfrac{\angle B-\angle A}{2},$ and in a similar manner we obtain $\angle OB_1A_2=\dfrac{\angle B-\angle A}{2},$ hence $\angle OA_1B_2=\angle OB_1A_2$. Thus, triangles $OA_1B_2$ and $OA_2B_1$ are equal, as $OA_1=OB_1$ and $A_1B_2=BC_1=AC_1=A_2B_1$ and $\angle OB_1A_2=\angle OA_1B_2$. Hence, $OA_2=OB_2$. Similarly, $OB_2=OC_2$, and so $O$ is the circumcenter of triangle $A_2B_2C_2,$ as desired.

Let $(ABC)$ be the unit circle and $a=x^2, b=y^2, c=z^2$ $$|x^2|=|x|\cdot|x|=1\implies|x|=1$$$$a_1=-yz,b_1=-xz,c_1=-xy$$$$a_2=-x^2-xy-xz=-x(x+y+z)$$$$|a_2|=|x|\cdot|x+y+z|=|x+y+z|$$similarly, $$|a_2|=|b_2|=|c_2|=|x+y+z|$$so their circumcenters are same we are done $\blacksquare$

(sry for long solution + no diagram) The idea is to shift our reference triangle from $ABC$ to $A_1B_1C_1$. Since $A_1$ is the arc midpoint of $\widehat{BC}$, it follows from angle chasing that $A$ is the intersection of the altitude from $A_1$ to $B_1C_1$ with $k$. The same fact applies for $B$ and $C$. So now the problem is as follows: $A_1B_1C_1$ is a triangle with circumcircle $k$. Let $A$ be the intersection of the foot from $A_1$ to $B_1C_1$ with $k$, and define $B$ and $C$ similarly. Then $A_2$, $B_2$, $C_2$ are just the reflections of $A$, $B$, $C$ over the midpoints of the opposite sides of our reference triangle $A_1B_1C_1$. We want to show that the circumcenter of $A_1B_1C_1$ is the same as the circumcenter of $A_2B_2C_2$. Now, let $O$ be the center of $k$, and let $H$ be the orthocenter of $A_1B_1C_1$. Then consider the antipode $A'$ of $A_1$ with respect to $k$. By the reflecting the orthocenter lemma, $A$ is the reflection of $H$ over $B_1C_1$ and $A'$ is the reflection of $H$ over the midpoint of $B_1C_1$. It follows that $HAA'A_2$ is a rectangle. Therefore, $A_2$ is the reflection of $H$ over the perpendicular bisector of $AA'$. Note that this is the perpendicular bisector of $B_1C_1$, so $O$ lies on this bisector. This implies $OH = OA_2$. By the same logic, we see that $OH = OA_2 = OB_2 = OC_2$, so $A_2$, $B_2$, $C_2$ lie on a circle with center $O$.

Proposed by Ivan Tagarev, Bulgaria (Pitagar/IvoBucata)