Let the table’s center be A and its radius be r. ABC is a right triangle, so (r – 8)² + (r – 9)² = r², or r² – 34r + 145 = 0, or (r – 5) × (r – 29) = 0. So r is 5 or 29, and the diameter is 10 or 58 inches — one solution is for a point on the table’s inside edge (shown), the other for a point on the outside edge.

From Steven Jerome Bryant et al., Nonroutine Problems in Algebra, Geometry, and Trigonometry, 1965.