Showing that a constructed point is the circumcenter of a given triangle For reference (exact copy of the question):

In the figure shown, what notable dot is "$E$" for triangle $ABC$? (answer: circumcenter)

Original figure:




My progress: The original figure does not make it clear whether the centers of the circles are coincident or whether points $A$ and $C$ are tangent to the smaller circle.
So I made this figure that seems to me to be the correct one.
Based on it I can say with certainty that:
$\angle E=90^\circ\\ EA=EC \therefore \triangle EAC(\text{isosceles})\implies \angle C=\angle A = 45^o$
So, $AE$ isn't a angle bisector and therefore $E$ isn't incenter
$AE$ is not perpendicular to $BC$ therefore $E$ is not orthocenter
The prolongation of $AE$ is not midpoint of $BC$ so $E$ is not barycenter
Need to show that $EA=EB$...


$\small \angle EAC=\angle AOC=\angle ECA=\angle CJA=45^\circ$ (alternate segment theorem)
Also $\small \angle BAC=\angle CJO$ and $\small \angle BCA=\angle AOJ$. (exterior angles of cyclic quadrilateral $\small AOJC$)
Say $\small \angle AJO=\alpha$ and $\small \angle COJ=\beta$. Thus $\small \angle BAE=\alpha$ and $\small \angle BCE=\beta$.
Considering $\small \triangle AOJ$, $\small \alpha+\beta+45^\circ=90^\circ\implies\alpha+\beta=45^\circ$
Considering $\small \triangle ABC$, $\small \alpha+\beta+45^\circ+45^\circ+\angle ABC=180^\circ$ $\small \implies\angle ABC=45^\circ$
Already found $\small \angle AEC=90^\circ$, $\small \angle ABC=\frac12 \angle AEC$. Also $\small AE=EC$. Therefore $\small E$ is the circumcentre of $\small \triangle ABC$.
