Question

Let $\mathcal{C}$ be the set of all the circles in a plane. If \[ \mathcal{R} = \{(C_1, C_2) \in \mathcal{C} \times \mathcal{C} \mid C_1 \text{ and } C_2 \text{ intersect}\} , \] then which of the following statements is TRUE?

• A. $\mathcal{R}$ is reflexive and symmetric but not transitive.
• B. $\mathcal{R}$ is reflexive and transitive but not symmetric.
• C. $\mathcal{R}$ is symmetric and transitive but not reflexive.
• D. $\mathcal{R}$ is not a relation.

Hint:

Geometric relations involving concepts like "intersection," "touches," or "is perpendicular to" are typically symmetric but rarely transitive.
Constructing a simple linear arrangement of objects (like three circles in a row) is an easy way to verify transitivity.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We evaluate the properties of the "intersection" relation on the set of circles.

Step 2: Detailed Explanation:
$\bullet$ Reflexive: Does every circle intersect itself? Yes, a circle shares all its points with itself. $(C_{1}, C_{1}) \in \mathcal{R}$. True.
$\bullet$ Symmetric: If $C_{1}$ intersects $C_{2}$, does $C_{2}$ intersect $C_{1}$? Yes, the property of sharing a point is mutual. True.
$\bullet$ Transitive: If $C_{1}$ intersects $C_{2}$, and $C_{2}$ intersects $C_{3}$, does $C_{1}$ necessarily intersect $C_{3}$?
Consider three circles in a row: $C_{1}$ on the left, $C_{2}$ in the middle overlapping with both, and $C_{3}$ on the right. $C_{1}$ and $C_{3}$ can be far apart and not touch each other at all. Thus, it is not transitive.

Step 3: Final Answer:
The relation is reflexive, symmetric, but not transitive.
This matches option (A).