From symmetry, $ D $ is the midpoint of $ BC $. Let the circle intersect $ BC $ again at $ X $ and $ BE $ again at $ Y $. The height of triangle $ ABC $ from $ A $ is $ \sqrt{26^2-10^2}=24 $ (from Pythagorean Theorem on $ \triangle ADB $), so the area of $ \triangle ABC $ is $ \left(\frac{1}{2}\right)(20)(24)=240 $. Therefore, since $ BE $ is an altitude, $ \left(\frac{1}{2}\right)(BE)(26)=240 $, so $ BE=\frac{240}{13} $. From Pythagorean Theorem on $ \triangle BEC $, $ EC=\sqrt{20^2-\left(\frac{240}{13}\right)^2}=\frac{100}{13} $. Now, using Power of a Point from $ C $, $ \left(\frac{100}{13}\right)^2=(10)(CX) $, so $ CX=\frac{1000}{169} $, and $ XD=10-CX=\frac{690}{169} $. From Power of a Point from $ B $, $ (BD)(BX)=(BY)(BE) $, so $ BY=\frac{BD\cdot BX}{BE}=\frac{(10)\left(10+\frac{690}{169}\right)}{\frac{240}{13}}=\frac{595}{78} $. Now $ YE=BE-BY=\frac{65}{6} $. Since the circle is tangent to $ AC $ at $ E $, the center lies on the perpendicular to $ AC $ at $ E $, which is $ YE $. Therefore, $ YE $ is the diameter, and so the radius is $ \frac{\frac{65}{6}}{2}=\boxed{\frac{65}{12}} $.

From symmetry, $D$ is the midpoint of $BC$. In Right $\triangle BCE, D$ is the midpoint of the hypotenuse, so $\angle BED=\angle EBC$ and $DE=BD=20/2=10$. Since $BE$ is perpendicular to $AC$, the center $O$, of the required circle, lies on $BE$. Let the altitude of isosceles $\triangle ODE$ from $O$ cut $DE$ at $X$. From symmetry, $EX=DE/2=5$. Also, Right $\triangle OEX$ is similar to Right $\triangle CAD$ because $\angle OEX=\angle BED=\angle EBC = \angle CAD$. By Pythagorean Theorem, $AD=24$. So the radius $OE=(CA/AD)(EX)=(26/24)(5)=65/12$.