Two equal circles are located so that each of them passes through the center of the other.

Two equal circles are located so that each of them passes through the center of the other. At what angle is their common chord visible from the center of each circle?

Let’s build the radii of the circles OA, O1A, OВ, O1B. The distance between the centers of the circles is equal to their radius, OO1 = R. Then the triangles OO1 and OВO1 are equilateral with internal angles 60.

Angle ABO = AOO1 + BOO1 = 60 + 60 = 120.

Angle AOB is the central angle that rests on the arc AB, then the chord AB is visible from the center of the circle at angles 120.

Similarly, the angle ABO1 = AO1O + BO1O = 60 + 60 = 120.

From point O1, the chord is also visible at an angle of 120.

Answer: The common chord AB is seen at an angle of 120.