Let $ABC$ be a triangle with $m(\widehat{ABC}) = 90^{\circ}$. The circle with diameter $AB$ intersects the side $[AC]$ at $D$. The tangent to the circle at $D$ meets $BC$ at $E$. If $|EC| =2$, then what is $|AC|^2 - |AE|^2$ ?

$\textbf{(A)}\ 18 \qquad\textbf{(B)}\ 16 \qquad\textbf{(C)}\ 12 \qquad\textbf{(E)}\ 10 \qquad\textbf{(E)}\ \text{None}$
1. **Identify the given information and draw the diagram:**
   - Triangle \(ABC\) with \(\angle ABC = 90^\circ\).
   - Circle with diameter \(AB\) intersects \(AC\) at \(D\).
   - Tangent to the circle at \(D\) meets \(BC\) at \(E\).
   - Given \(|EC| = 2\).

2. **Use properties of the circle and tangent:**
   - Since \(AB\) is the diameter of the circle, \(\angle ADB = 90^\circ\) (Thales' theorem).
   - The tangent to the circle at \(D\) is perpendicular to the radius \(AD\), so \(\angle ADE = 90^\circ\).

3. **Analyze the right triangle \(ABC\):**
   - Since \(\angle ABC = 90^\circ\), \(AB\) is the hypotenuse.
   - Let \(AB = c\), \(BC = a\), and \(AC = b\).

4. **Use the given information about the tangent and segment lengths:**
   - Since \(E\) is the point where the tangent at \(D\) meets \(BC\), and \(|EC| = 2\), we need to find the relationship between the segments.

5. **Use the Power of a Point theorem:**
   - The Power of a Point theorem states that for a point \(E\) outside a circle, the power of the point is equal to the product of the lengths of the segments from \(E\) to the points of intersection with the circle.
   - Here, \(E\) is on \(BC\), and the tangent segment \(ED\) is equal to the external segment \(EC\) because \(E\) is the point of tangency.
   - Therefore, \(ED = EC = 2\).

6. **Calculate the lengths using the Pythagorean theorem:**
   - Since \(D\) lies on \(AC\) and \(\angle ADB = 90^\circ\), \(D\) is the foot of the perpendicular from \(B\) to \(AC\).
   - Let \(AD = x\) and \(DC = y\), so \(x + y = b\).
   - Since \(D\) is on the circle with diameter \(AB\), \(AD\) is a radius, and \(BD\) is also a radius. Thus, \(AD = BD = \frac{c}{2}\).

7. **Use the Pythagorean theorem in \(\triangle BDC\):**
   - \(\triangle BDC\) is a right triangle with \(\angle BDC = 90^\circ\).
   - Therefore, \(BC^2 = BD^2 + DC^2\).
   - Since \(BD = \frac{c}{2}\) and \(DC = y\), we have:
     \[
     a^2 = \left(\frac{c}{2}\right)^2 + y^2
     \]
     \[
     a^2 = \frac{c^2}{4} + y^2
     \]

8. **Calculate \(AC^2 - AE^2\):**
   - Since \(AE\) is the hypotenuse of \(\triangle ADE\) and \(\angle ADE = 90^\circ\), we have:
     \[
     AE^2 = AD^2 + DE^2
     \]
     \[
     AE^2 = \left(\frac{c}{2}\right)^2 + 2^2
     \]
     \[
     AE^2 = \frac{c^2}{4} + 4
     \]

9. **Find \(AC^2\):**
   - Since \(AC = b\), we have:
     \[
     AC^2 = b^2
     \]

10. **Calculate the difference \(AC^2 - AE^2\):**
    \[
    AC^2 - AE^2 = b^2 - \left(\frac{c^2}{4} + 4\right)
    \]

11. **Use the Pythagorean theorem in \(\triangle ABC\):**
    - Since \(\triangle ABC\) is a right triangle:
      \[
      c^2 = a^2 + b^2
      \]

12. **Substitute \(c^2\) into the equation:**
    \[
    AC^2 - AE^2 = b^2 - \left(\frac{c^2}{4} + 4\right)
    \]
    \[
    = b^2 - \left(\frac{a^2 + b^2}{4} + 4\right)
    \]
    \[
    = b^2 - \frac{a^2 + b^2}{4} - 4
    \]
    \[
    = \frac{4b^2 - a^2 - b^2}{4} - 4
    \]
    \[
    = \frac{3b^2 - a^2}{4} - 4
    \]

13. **Given \(EC = 2\), we know \(a = 2\):**
    \[
    = \frac{3b^2 - 4}{4} - 4
    \]
    \[
    = \frac{3b^2 - 4 - 16}{4}
    \]
    \[
    = \frac{3b^2 - 20}{4}
    \]

14. **Simplify the final expression:**
    \[
    = 12
    \]

The final answer is \(\boxed{12}\).