Let $ ABC$ be a triangle with $ \angle BAC \equal{} 90^\circ$.  A circle is tangent to the sides $ AB$ and $ AC$ at $ X$ and $ Y$ respectively, such that the points on the circle diametrically opposite $ X$ and $ Y$ both lie on the side $ BC$. Given that $ AB \equal{} 6$, find the area of the portion of the circle that lies outside the triangle.

[asy]import olympiad;
import math;
import graph;

unitsize(20mm);
defaultpen(fontsize(8pt));

pair A = (0,0);
pair B = A + right;
pair C = A + up;

pair O = (1/3, 1/3);

pair Xprime = (1/3,2/3);
pair Yprime = (2/3,1/3);

fill(Arc(O,1/3,0,90)--Xprime--Yprime--cycle,0.7*white);

draw(A--B--C--cycle);
draw(Circle(O, 1/3));
draw((0,1/3)--(2/3,1/3));
draw((1/3,0)--(1/3,2/3));

label("$A$",A, SW);
label("$B$",B,  down);
label("$C$",C,  left);
label("$X$",(1/3,0),  down);
label("$Y$",(0,1/3),  left);[/asy]
1. Given that $\angle BAC = 90^\circ$, triangle $ABC$ is a right triangle with $A$ as the right angle. The circle is tangent to $AB$ at $X$ and to $AC$ at $Y$, and the points diametrically opposite $X$ and $Y$ lie on $BC$.
2. Let $AX = AY = r$, where $r$ is the radius of the circle. Since the circle is tangent to $AB$ and $AC$, the center of the circle, $O$, lies on the angle bisector of $\angle BAC$. Therefore, $O$ is at $(r, r)$.
3. The points diametrically opposite $X$ and $Y$ are $X'$ and $Y'$, respectively. Since $X'$ and $Y'$ lie on $BC$, the diameter of the circle is perpendicular to $BC$. Thus, the diameter of the circle is $2r$.
4. Since $AB = 6$, and $AX = r$, we have $AB = AX + XB = r + XB$. Therefore, $XB = 6 - r$.
5. Similarly, $AC = 6$, and $AY = r$, we have $AC = AY + YC = r + YC$. Therefore, $YC = 6 - r$.
6. Since $X'$ and $Y'$ lie on $BC$, the length of $BC$ is equal to the diameter of the circle, which is $2r$. Therefore, $BC = 2r$.
7. Using the Pythagorean theorem in $\triangle ABC$, we have:
   \[
   AB^2 + AC^2 = BC^2
   \]
   Substituting the given values:
   \[
   6^2 + 6^2 = (2r)^2
   \]
   \[
   36 + 36 = 4r^2
   \]
   \[
   72 = 4r^2
   \]
   \[
   r^2 = 18
   \]
   \[
   r = 3\sqrt{2}
   \]
8. The area of the circle is:
   \[
   \pi r^2 = \pi (3\sqrt{2})^2 = 18\pi
   \]
9. The area of $\triangle ABC$ is:
   \[
   \frac{1}{2} \times AB \times AC = \frac{1}{2} \times 6 \times 6 = 18
   \]
10. The area of the portion of the circle that lies outside the triangle is the area of the circle minus the area of the triangle:
    \[
    18\pi - 18
    \]

The final answer is $\boxed{18\pi - 18}$