13. As shown in Figure 3, two perpendicular lines intersect at the center of a circle with a radius of 1, dividing the circle into four parts. Another smaller circle is tangent to one of these parts. The radius of the smaller circle is ( ).
(A) $\frac{1}{3}$
(B) $\frac{2}{5}$
(C) $\sqrt{2}-1$
(D) $\frac{1}{2}$
(E) $2-\sqrt{2}$
13. C.

Let the radius of the small circle be $x$, then we have $x^{2}+x^{2}=(1-x)^{2}$, which simplifies to $x^{2}+2 x-1=0$.
Solving for $x$ gives $x= \pm \sqrt{2}-1$.
Since $x>0$, we have $x=\sqrt{2}-1$.