To solve the problem, we analyze the sets (A) and (B) and their intersection:

Step 1: Define the sets

• Set (A = {(x,y) \mid x^2 + y^2 = 1}): This is the unit circle (boundary of the unit disk).
• Set (B = {(x,y) \mid x^2 + y^2 \leq 1}): This is the unit disk (including its boundary).

Step 2: Find the intersection (A \cap B)

An element ((x,y)) is in (A \cap B) if it belongs to both (A) and (B).

For any ((x,y) \in A), (x^2 + y^2 = 1), which satisfies (x^2 + y^2 \leq 1) (since (1 \leq 1)). Thus, every element of (A) is in (B).

Conversely, elements in (B) but not in (A) have (x^2 + y^2 < 1), so they are not in (A).

Answer: (\boxed{{(x,y) \mid x^2 + y^2 = 1}}) (or simply the unit circle, but in set notation as above).

But if the expected answer is a concise form, it's the unit circle, so the boxed answer can be written as:

(\boxed{x^2 + y^2 = 1}) (since the question asks for the intersection, which is the set defined by this equation).

(\boxed{x^2 + y^2 = 1})

作者声明:本文包含人工智能生成内容。