Let $ABC$ be a triangle with circumradius $1$. If the center of the circle passing through $A$, $C$, and the orthocenter of $\triangle ABC$ lies on the circumcircle of $\triangle ABC$, what is $|AC|$?

$ 
\textbf{(A)}\ 2
\qquad\textbf{(B)}\ 3
\qquad\textbf{(C)}\ \dfrac 32
\qquad\textbf{(D)}\ \sqrt 2 
\qquad\textbf{(E)}\ \sqrt 3
$
1. Let $ABC$ be a triangle with circumradius $R = 1$. The orthocenter of $\triangle ABC$ is denoted by $H$, and the circumcenter is denoted by $O$. We are given that the center of the circle passing through $A$, $C$, and $H$ lies on the circumcircle of $\triangle ABC$.

2. Let $K$ be the center of the circle passing through $A$, $C$, and $H$. Since $K$ lies on the circumcircle of $\triangle ABC$, $K$ must be the midpoint of the arc $AC$ that does not contain $B$. This is because the center of the circle passing through $A$, $C$, and $H$ must lie on the perpendicular bisector of $AC$ and also on the circumcircle of $\triangle ABC$.

3. Since $K$ is the midpoint of the arc $AC$, it is equidistant from $A$ and $C$. Therefore, $KA = KC$. Also, $K$ lies on the perpendicular bisector of $AH$ and $CH$.

4. Consider the triangle $AHC$. Since $K$ is the circumcenter of $\triangle AHC$, the radius of the circumcircle of $\triangle AHC$ is $KA = KC$. Let $M$ be the midpoint of $BC$. The length $AH$ can be expressed as $AH = 2R \cos A = 2 \cos A$ because $R = 1$.

5. Since $K$ is the midpoint of the arc $AC$, the angle $\angle AKC = 180^\circ - \angle AOC = 180^\circ - 2B$. Therefore, $\angle AKC = 2(90^\circ - B)$. 

6. Using the Law of Cosines in $\triangle AKC$, we have:
   \[
   AC^2 = KA^2 + KC^2 - 2 \cdot KA \cdot KC \cdot \cos(\angle AKC)
   \]
   Since $KA = KC = 1$ (radius of the circumcircle of $\triangle ABC$), we get:
   \[
   AC^2 = 1^2 + 1^2 - 2 \cdot 1 \cdot 1 \cdot \cos(2(90^\circ - B))
   \]
   \[
   AC^2 = 2 - 2 \cos(180^\circ - 2B)
   \]
   \[
   AC^2 = 2 - 2(-\cos 2B)
   \]
   \[
   AC^2 = 2 + 2 \cos 2B
   \]

7. Since $K$ lies on the circumcircle of $\triangle ABC$, $B = 60^\circ$. Therefore, $\cos 2B = \cos 120^\circ = -\frac{1}{2}$.

8. Substituting $\cos 120^\circ = -\frac{1}{2}$ into the equation, we get:
   \[
   AC^2 = 2 + 2 \left(-\frac{1}{2}\right)
   \]
   \[
   AC^2 = 2 - 1
   \]
   \[
   AC^2 = 1
   \]
   \[
   AC = \sqrt{1} = 1
   \]

However, this contradicts the given options. Let's re-evaluate the problem with the correct approach.

9. Given $B = 60^\circ$, we use the fact that $AC = 2R \sin B$. Since $R = 1$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$, we have:
   \[
   AC = 2 \cdot 1 \cdot \sin 60^\circ = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}
   \]

The final answer is $\boxed{\sqrt{3}}$