26. As shown in the figure, in tetrahedron $ABCD$, $CD$ is the common perpendicular to the opposite edges $AD$ and $BC$. Given that $CD=a$, $AB=b$, and the angle between $AD$ and $BC$ is $30^{\circ}$. Find the radius of the circumscribed sphere of tetrahedron $ABCD$.
26. Draw $CE \parallel DA$, then $\angle ECB=30^{\circ}$. Take $CE=DA$, connect $AE$, then $AE \parallel CD$. Since $CD$ is the common perpendicular of $AD$ and $BC$, hence $CD \perp BC, CD \perp AD, \therefore \quad CD \perp$ plane $BCE, \therefore \quad AE \perp$ plane $BCE$.

Let $O$ be the center of the circumscribed sphere of tetrahedron $ABCD$, take the midpoints $F, G$ of $CD$ and $AE$. Since $OD=OC, \therefore OF \perp CD, \therefore$ $AE \perp OF$. Also, $FG \perp AE$, if $F, O, G$ are not collinear, $AE \perp$ plane $FOG$, at this time $OG \perp AE$; if $F, O, G$ are collinear, obviously $OG \perp AE, \therefore OA=OE$. Since $OB, OC, OE$ are equal, their projections in plane $BCE$ are equal, hence $O$'s projection $O_1$ in plane $BCE$ is the circumcenter of $\triangle BCE$.

In the right triangle $\triangle AEB$, $AB=\sqrt{b^2-a^2}$. In $\triangle BCE$, by the sine rule $O_1C=\frac{1}{2} \frac{EB}{\sin 30^{\circ}}=\sqrt{b^2-a^2}$. Since $OFCO_1$ is a rectangle, hence $OO_1=\frac{a}{2}$. In the right triangle $\triangle OCO_1$, by the Pythagorean theorem, the radius of the circumscribed sphere $R=OC=$ $\sqrt{\left(\frac{a}{2}\right)^2+\left(b^2-a^2\right)}=\frac{1}{2} \sqrt{4b^2-3a^2}$.