The cyclic octagon $ABCDEFGH$ has sides $a,a,a,a,b,b,b,b$ respectively. Find the radius of the circle that circumscribes $ABCDEFGH.$
1. **Rearrange the vertices**: We can rearrange the vertices of the octagon so that the side lengths alternate between \(a\) and \(b\). This does not change the circumradius of the octagon. Thus, we can consider the octagon with vertices \(P, Q, R, S, T, U, V, W\) such that \(PQ = a\), \(QR = b\), \(RS = a\), and so on.

2. **Symmetry and angle calculation**: By symmetry, the angle \(\angle POR\) is \(90^\circ\). This implies that \(\angle PQR = 135^\circ\) because the internal angle of a regular octagon is \(135^\circ\).

3. **Law of cosines**: Apply the Law of Cosines to \(\triangle PQR\):
   \[
   PR^2 = PQ^2 + QR^2 - 2 \cdot PQ \cdot QR \cdot \cos(135^\circ)
   \]
   Since \(\cos(135^\circ) = -\frac{1}{\sqrt{2}}\), we have:
   \[
   PR^2 = a^2 + b^2 - 2ab \left(-\frac{1}{\sqrt{2}}\right) = a^2 + b^2 + \sqrt{2}ab
   \]
   Therefore,
   \[
   PR = \sqrt{a^2 + b^2 + \sqrt{2}ab}
   \]

4. **Right isosceles triangle**: The triangle \(\triangle POR\) is a right isosceles triangle with \(PO = RO = r\). Since \(\angle POR = 90^\circ\), we can use the properties of a right isosceles triangle to find \(r\):
   \[
   PR = r\sqrt{2}
   \]
   Solving for \(r\), we get:
   \[
   r = \frac{PR}{\sqrt{2}} = \frac{\sqrt{a^2 + b^2 + \sqrt{2}ab}}{\sqrt{2}} = \sqrt{\frac{a^2 + b^2 + \sqrt{2}ab}{2}}
   \]

Thus, the radius of the circle that circumscribes the octagon is:
\[
\boxed{\sqrt{\frac{a^2 + b^2 + \sqrt{2}ab}{2}}}
\]