It is easy to get: △ABO and △DCE are isosceles right triangles, which can prove that the midpoint between AB side and DC side is a point, so point E moves on straight line y=-x, so comparing the positional relationship between ○o and straight line y=x+b is to compare the height of AB side of △ABO and DC side of △DCE, and because DC=r is a circle O, the height of AB side is 65433. So compare the relationship between DC and 1/2AB, and because AO=BO=b and AB=√2b, the height on the side of AB is √2b/2, and DC=√2√(b? - 16)。

When the straight line is tangent to o: √2b/2=√2√(b? - 16)

B= 8/3 root number three because B > 4, so when B = 8/3 root number three lines separate or intersect with O: because 8/3 root number three ≈4.6, B > 4, so b=5.

When b=5, √2b/2≈3.5, √2√(b? - 16)=3√2 ≈4.2

Because when b=5, DC is higher than the height on the side of AB, and when the straight line intersects with O, B > 8/3. When the straight line deviates from o, 4 < b < 8/3 root number three.

The above was written on the test paper when I was in the senior high school entrance examination. All rights reserved, plagiarism will be investigated. If there are similarities, it is purely coincidental.