In the figure given, the tangent to the circular ring in the plane of the ring is shown. Let, the tangent be represented by XY and diameter be represented by AB. Using the theorem of parallel axes, we have IXY = IAB + M (R)2       =  i.e., IXY = Therefore, the moment of inertia of ring about tangent in the plane of the ring is 3/2 MR2.