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?
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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.
$\mathcal{R}$ is reflexive and symmetric but not transitive.
$\mathcal{R}$ is reflexive and transitive but not symmetric.
$\mathcal{R}$ is symmetric and transitive but not reflexive.
$\mathcal{R}$ is not a relation.
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The Correct Option isA
Solution and Explanation
Step 1: Understanding the Question:
We are given a relation $\mathcal{R}$ defined on the set of all circles in a plane, $\mathcal{C}$.
Two circles $C_1$ and $C_2$ are related if they intersect.
We need to test this relation for reflexivity, symmetry, and transitivity. Step 2: Key Formula or Approach: • Reflexivity: A relation is reflexive if $(C_1, C_1) \in \mathcal{R}$ for every $C_1 \in \mathcal{C}$.
• Symmetry: A relation is symmetric if $(C_1, C_2) \in \mathcal{R} \implies (C_2, C_1) \in \mathcal{R}$.
• Transitivity: A relation is transitive if $(C_1, C_2) \in \mathcal{R}$ and $(C_2, C_3) \in \mathcal{R} \implies (C_1, C_3) \in \mathcal{R}$. Step 3: Detailed Explanation:
• Reflexivity:
Let $C_1$ be any circle in the plane.
Every circle completely overlaps with itself, which means it shares all of its points with itself.
Thus, $C_1$ intersects $C_1$.
So, $(C_1, C_1) \in \mathcal{R}$ for all $C_1 \in \mathcal{C}$.
Therefore, the relation $\mathcal{R}$ is reflexive.
• Symmetry:
Let $C_1, C_2 \in \mathcal{C}$ such that $(C_1, C_2) \in \mathcal{R}$.
This means $C_1$ and $C_2$ intersect, sharing at least one common point.
If $C_1$ intersects $C_2$, then $C_2$ must also intersect $C_1$.
Thus, $(C_2, C_1) \in \mathcal{R}$.
Therefore, the relation $\mathcal{R}$ is symmetric.
• Transitivity:
Let $C_1, C_2, C_3 \in \mathcal{C}$ such that $(C_1, C_2) \in \mathcal{R}$ and $(C_2, C_3) \in \mathcal{R}$.
This means $C_1$ intersects $C_2$, and $C_2$ intersects $C_3$.
However, this does not guarantee that $C_1$ intersects $C_3$.
For example, let $C_1$ be a circle centered at $(0,0)$ with radius $1$.
Let $C_2$ be a circle centered at $(1.5,0)$ with radius $1$.
These two circles intersect because the distance between their centers ($1.5$) is less than the sum of their radii ($2$).
Now, let $C_3$ be a circle centered at $(3,0)$ with radius $1$.
$C_2$ and $C_3$ intersect because the distance between their centers ($1.5$) is less than $2$.
But the distance between the centers of $C_1$ and $C_3$ is $3$, which is greater than the sum of their radii ($2$).
Therefore, $C_1$ and $C_3$ do not intersect.
So, $(C_1, C_2) \in \mathcal{R}$ and $(C_2, C_3) \in \mathcal{R}$, but $(C_1, C_3) \notin \mathcal{R}$.
Thus, the relation $\mathcal{R}$ is not transitive. Step 4: Final Answer:
The relation $\mathcal{R}$ is reflexive and symmetric but not transitive.
