If a relation R is reflexive and circular then it is symmetric : True
Proof : Assume $a\text{R}b.$ Then since $R$ is reflexive, we have $b\text{R}b.$ Since $R$ is circular, so, $a\text{R}b, b\text{R}b$ will mean that we have $b\text{R}a.$ So, $R$ is symmetric.
If R is reflexive and circular then it is transitive : True
Proof : Assume $a\text{R}b, b\text{R}c.$ Since $R$ is circular, so, $c\text{R}a,$ and since $R$ is symmetric(we proved above) so $a\text{R}c$ so $R$ is transitive.
So, option C is correct.
Option B is false. For counter example, take a set $A = \{a,b,c\},$ define relation $R$ on $A$ as follows : $R = \{ (a,a) \}, R$ is symmetric and circular but not equivalence relation.
Option A is false. For counter example, take a set $A = \{a,b,c\},$ define relation $R$ on $A$ as follows: $R = \{ (a,a), (b,b),(c,c), (a,b),(b,a), (a,c),(c,a) \},$ $R$ is symmetric and reflexive but not transitive so not equivalence relation.
Option D is false. For counter example, take a set $A = \{a,b,c\},$ define relation R on A as follows: $R = \{ (a,a) \},$ $R$ is transitive and circular but not equivalence relation.
Some more variations :
1. Converse of Statement in option C is also true. i.e.
Theorem : If R is an equivalence relation then R is reflexive and circular.
Proof :
Reflexive: As, the relation $R$ is an equivalence relation. So, reflexivity is the property of an equivalence relation. Hence, $R$ is reflexive.
Circular: Let $(a, b) \in R$ and $(b, c) ∈ R$
$⇒ (a, c) ∈ R$ (∵ R is transitive)
$⇒ (c, a) ∈ R$ (∵ R is symmetric)
Thus, $R$ is Circular.
So, we can say that
“A relation S is reflexive and circular if and only if S is an equivalence relation.”
2. If a relation $R$ is transitive and circular then it is symmetric : False.
3. If a relation $R$ is transitive and circular then it is reflexive : False.
Counter example(for both above statements) : $R = \{ (a,b) \}$
PS : Similarly you can try to prove or disprove more similar statements and their converses.