A, O and C are collinear. Since a radius is perpendicular to its tangent, \(OB\perp EB\); the parallel diameter AC gives \(\angle AOB=90^\circ\). Therefore \(\angle DOB=44^\circ+90^\circ=134^\circ\).

Quadrilateral ODEB has right angles at D and B, so \(\angle DEB=360^\circ-90^\circ-90^\circ-134^\circ=46^\circ\). Right triangles EOD and EOB are congruent (common hypotenuse EO and equal radii), so EO bisects this angle. Thus \(\boxed{x=23^\circ}\).