An [i]annulus[/i] is the region between two concentric circles. The concentric circles in the ﬁgure have radii $ b$ and $ c$, with $ b > c$. Let $ \overline{OX}$ be a radius of the larger circle, let $ \overline{XZ}$ be tangent to the smaller circle at $ Z$, and let $ \overline{OY}$ be the radius of the larger circle that contains $ Z$. Let $ a \equal{} XZ$, $ d \equal{} YZ$, and $ e \equal{} XY$. What is the area of the annulus?

$ \textbf{(A)}\ \pi a^2 \qquad \textbf{(B)}\ \pi b^2 \qquad \textbf{(C)}\ \pi c^2 \qquad \textbf{(D)}\ \pi d^2 \qquad \textbf{(E)}\ \pi e^2$
[asy]unitsize(1.4cm);
defaultpen(linewidth(.8pt));
dotfactor=3;

real r1=1.0, r2=1.8;
pair O=(0,0), Z=r1*dir(90), Y=r2*dir(90);
pair X=intersectionpoints(Z--(Z.x+100,Z.y), Circle(O,r2))[0];
pair[] points={X,O,Y,Z};

filldraw(Circle(O,r2),mediumgray,black);
filldraw(Circle(O,r1),white,black);

dot(points);
draw(X--Y--O--cycle--Z);

label("$O$",O,SSW,fontsize(10pt));
label("$Z$",Z,SW,fontsize(10pt));
label("$Y$",Y,N,fontsize(10pt));
label("$X$",X,NE,fontsize(10pt));

defaultpen(fontsize(8pt));

label("$c$",midpoint(O--Z),W);
label("$d$",midpoint(Z--Y),W);
label("$e$",midpoint(X--Y),NE);
label("$a$",midpoint(X--Z),N);
label("$b$",midpoint(O--X),SE);[/asy]
1. The area of the annulus is the difference between the area of the larger circle and the area of the smaller circle. The formula for the area of a circle is given by:
   \[
   \text{Area} = \pi r^2
   \]
   Therefore, the area of the larger circle with radius \( b \) is:
   \[
   \pi b^2
   \]
   and the area of the smaller circle with radius \( c \) is:
   \[
   \pi c^2
   \]

2. The area of the annulus is then:
   \[
   \text{Area of annulus} = \pi b^2 - \pi c^2 = \pi (b^2 - c^2)
   \]

3. Next, we need to express \( b^2 - c^2 \) in terms of \( a \). Consider the right triangle \( \triangle XOZ \), where \( \overline{XZ} \) is tangent to the smaller circle at \( Z \). Since \( \overline{XZ} \) is tangent to the smaller circle, \( \angle XZO \) is a right angle. By the Pythagorean theorem, we have:
   \[
   XZ^2 + OZ^2 = OX^2
   \]
   Substituting the given lengths, we get:
   \[
   a^2 + c^2 = b^2
   \]

4. Rearranging the equation to solve for \( b^2 - c^2 \), we get:
   \[
   b^2 - c^2 = a^2
   \]

5. Substituting this back into the expression for the area of the annulus, we get:
   \[
   \text{Area of annulus} = \pi (b^2 - c^2) = \pi a^2
   \]

The final answer is \(\boxed{\pi a^2}\).