ABCD is a square of unit area. A circle is tangent to two sides of ABCD and passes through exactly one of its vertices. The radius of the circle is: 

Suppose that $ABCD$ is a square, axes are along two sides of the sqaure. 

let $r$ be the radius of the circle, and passing through the point $C\left(1,1\right)$

The equation of any circle whose radius $r$ and touches both axes is $x^2+y^2-2rx-2ry+r^2=0$

And it passing through the point $\left(1,1\right)$

$\begin{array}{rcl}1+1-2r-2r+r^2& =& 0\\ r^2-4r+2& =& 0\\ r& =& \frac{4\pm \sqrt{16-8}}{2}\\ & =& \frac{4\pm 2\sqrt{2}}{2}\\ & =& 2\pm \sqrt{2}\end{array}$

Since the circle is passing throgh one vertex $\left(1,1\right)$, hence the radius must be less than 1

hene the radius is $\boxed{2-\sqrt{2}}$