Two variable chords AB and BC of a circle `x^(2)+y^(2)=a^(2)` are such that `AB=BC=a`. M and N are the midpoints of AB and BC, respectively, such that the line joining MN intersects the circles at P and Q, where P is closer to AB and O is the center of the circle.
`/_ OAB` is

Options

A. `15^(@)`
B. `30^(@)`
C. `45^(@)`
D. `60^(@)`

Answer

To solve the problem, we need to find the angle \( \angle OAB \) where \( O \) is the center of the circle defined by the equation \( x^2 + y^2 = a^2 \), and \( A \) and \( B \) are points on the circle such that the chord \( AB \) has a length equal to \( a \). The chord \( BC \) is also equal to \( a \).

### Step-by-Step Solution:

1. **Understand the Circle**:
The equation of the circle is given by \( x^2 + y^2 = a^2 \). This means the center \( O \) of the circle is at the origin \( (0, 0) \) and the radius \( r \) is \( a \).

**Hint**: Remember that the radius of the circle is the distance from the center to any point on the circle.

2. **Draw the Chords**:
Let's denote the points \( A \) and \( B \) on the circle such that the length of chord \( AB = a \). Similarly, denote point \( C \) such that \( BC = a \).

**Hint**: Visualizing the chords and the circle will help in understanding the relationships between the points.

3. **Midpoints**:
Let \( M \) be the midpoint of chord \( AB \) and \( N \) be the midpoint of chord \( BC \). The coordinates of \( M \) and \( N \) can be calculated if we know the coordinates of points \( A \), \( B \), and \( C \).

**Hint**: The midpoint of a segment can be found using the formula \( M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \).

4. **Triangles Involved**:
The triangle \( OAB \) is formed by the center \( O \) and the endpoints of the chord \( AB \). Since \( OA = OB = a \) (the radius) and \( AB = a \), triangle \( OAB \) is an equilateral triangle.

**Hint**: In an equilateral triangle, all sides are equal, and all angles are \( 60^\circ \).

5. **Finding the Angle**:
Since triangle \( OAB \) is equilateral, we can conclude that:
\[ \angle OAB = 60^\circ \]

**Hint**: Recall that in an equilateral triangle, each angle measures \( 60^\circ \).

### Final Answer:
Thus, the angle \( \angle OAB \) is \( 60^\circ \).