1. Understand the configuration

   Two equal masses, each of mass $m$, revolve in a circle due to their mutual gravitational attraction.

   Since the masses are equal, they rotate about their common center of mass, which lies exactly midway between them.

2. Radius of each particle's circular path

   The question says each particle moves in a circle of radius $r$.

   Therefore, the distance between the two particles is $2r.$

3. Gravitational force between the two masses

   The gravitational attraction between them is $F = \frac{Gm^2}{(2r)^2} = \frac{Gm^2}{4r^2}.$

4. Centripetal force required for each particle

   Each particle moves in a circle of radius $r$ with speed $v$, so the required centripetal force is $F_c = \frac{mv^2}{r}.$

5. Use gravitational force as centripetal force

   Since gravity provides the centripetal force, $\frac{mv^2}{r} = \frac{Gm^2}{4r^2}.$

6. Solve for $v$

   Cancel $m$ from both sides: $\frac{v^2}{r} = \frac{Gm}{4r^2}.$

   Multiply by $r$: $v^2 = \frac{Gm}{4r}.$

   Hence, $v = \sqrt{\frac{Gm}{4r}}.$

7. Match with the options

   This corresponds to Option A.

Verification with stored answer: Stored correct answer is A, which matches the derived result.