[asy]
size(150);
pair A=(0,0),B=(1,0),C=(0,1),D=(-1,0),E=(0,.5),F=(sqrt(2)/2,.25);
draw(circle(A,1)^^D--B);
draw(circle(E,.5)^^circle( F ,.25));
label("$A$", D, W);
label("$K$", A, S);
label("$B$", B, dir(0));
label("$L$", E, N);
label("$M$",shift(-.05,.05)*F);
//Credit to Klaus-Anton for the diagram[/asy]

In the adjoining figure, circle $\mathit{K}$ has diameter $\mathit{AB}$; cirlce $\mathit{L}$ is tangent to circle $\mathit{K}$ and to $\mathit{AB}$ at the center of circle $\mathit{K}$; and circle $\mathit{M}$ tangent to circle $\mathit{K}$, to circle $\mathit{L}$ and $\mathit{AB}$. The ratio of the area of circle $\mathit{K}$ to the area of circle $\mathit{M}$ is

$\textbf{(A) }12\qquad\textbf{(B) }14\qquad\textbf{(C) }16\qquad\textbf{(D) }18\qquad \textbf{(E) }\text{not an integer}$
1. **Scale the figure**: We scale the figure so that the radius of circle \( K \) is \( 2 \). This means the diameter \( AB \) of circle \( K \) is \( 4 \).

2. **Determine the radius of circle \( L \)**: Since circle \( L \) is tangent to circle \( K \) at the center of circle \( K \), and circle \( K \) has a diameter of \( 4 \), the radius of circle \( L \) is half of this diameter, which is \( 1 \).

3. **Set up the problem for circle \( M \)**: Let the radius of circle \( M \) be \( r \). Circle \( M \) is tangent to circle \( K \), circle \( L \), and the line \( AB \).

4. **Calculate distances**:
   - The distance from the center of circle \( L \) to the center of circle \( M \) is \( 1 + r \) (since they are tangent to each other).
   - The distance from the center of circle \( K \) to the center of circle \( M \) is \( 2 - r \) (since they are tangent to each other).

5. **Use the Pythagorean theorem**:
   - Let \( Q \) be the point where the line through the center of circle \( M \) perpendicular to \( LK \) meets \( LK \).
   - The distance \( LQ = 1 - r \) (since \( L \) and \( M \) are collinear and \( LQ \) is the vertical distance from \( L \) to \( M \)).
   - The distance \( LM = 1 + r \).

   By the Pythagorean theorem in \( \triangle LQM \):
   \[
   (1 + r)^2 = (1 - r)^2 + QM^2
   \]
   Simplifying:
   \[
   1 + 2r + r^2 = 1 - 2r + r^2 + QM^2
   \]
   \[
   4r = QM^2
   \]
   Thus:
   \[
   QM = \sqrt{4r} = 2\sqrt{r}
   \]

6. **Relate \( QM \) to \( KP \)**:
   - Since \( KP \) is the same as \( QM \) (both are vertical distances from the center of circle \( K \) to the tangent point on circle \( M \)):
   \[
   KP = 2\sqrt{r}
   \]

7. **Use the Pythagorean theorem in \( \triangle KPM \)**:
   - \( KM = 2 - r \)
   - \( MP = r \)
   - \( KP = 2\sqrt{r} \)

   By the Pythagorean theorem in \( \triangle KPM \):
   \[
   (2 - r)^2 + r^2 = (2\sqrt{r})^2
   \]
   Simplifying:
   \[
   4 - 4r + r^2 + r^2 = 4r
   \]
   \[
   4 - 4r + 2r^2 = 4r
   \]
   \[
   2r^2 - 8r + 4 = 0
   \]
   Solving this quadratic equation:
   \[
   r^2 - 4r + 2 = 0
   \]
   Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   r = \frac{4 \pm \sqrt{16 - 8}}{2} = \frac{4 \pm \sqrt{8}}{2} = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2}
   \]
   Since \( r \) must be positive and less than 2, we take \( r = 2 - \sqrt{2} \).

8. **Calculate the ratio of the areas**:
   - The area of circle \( K \) is \( \pi \times 2^2 = 4\pi \).
   - The area of circle \( M \) is \( \pi \times (2 - \sqrt{2})^2 \).

   Simplifying the area of circle \( M \):
   \[
   (2 - \sqrt{2})^2 = 4 - 4\sqrt{2} + 2 = 6 - 4\sqrt{2}
   \]
   Thus, the area of circle \( M \) is \( \pi (6 - 4\sqrt{2}) \).

   The ratio of the areas is:
   \[
   \frac{4\pi}{\pi (6 - 4\sqrt{2})} = \frac{4}{6 - 4\sqrt{2}}
   \]

   Rationalizing the denominator:
   \[
   \frac{4}{6 - 4\sqrt{2}} \times \frac{6 + 4\sqrt{2}}{6 + 4\sqrt{2}} = \frac{4(6 + 4\sqrt{2})}{(6 - 4\sqrt{2})(6 + 4\sqrt{2})}
   \]
   \[
   = \frac{24 + 16\sqrt{2}}{36 - (4\sqrt{2})^2} = \frac{24 + 16\sqrt{2}}{36 - 32} = \frac{24 + 16\sqrt{2}}{4} = 6 + 4\sqrt{2}
   \]

   Since this is not an integer, the correct answer is \( \text{not an integer} \).

The final answer is \( \boxed{ \text{not an integer} } \).