O is the centre of a circle. There is a point P in the region of the circle. If PA = PB = PC where A, B and C are points on the circumference of the circle, then OP must be equal to :

Options

A. `(PA + PB + PC)/(3)`
B. `(PA + PB + PC)/(2)`
C. `(AB + BC)/(2)`
D. zero

Answer

To solve the problem, we need to analyze the given information step by step.

### Step-by-Step Solution:

1. **Identify the Elements**: We have a circle with center O and points A, B, and C on the circumference. There is a point P inside the circle such that the distances from P to A, B, and C are equal (PA = PB = PC).

2. **Understanding the Condition**: The condition PA = PB = PC indicates that point P is equidistant from points A, B, and C. This can only occur if point P is located at the center of the circle.

3. **Distance from Center to Circumference**: In a circle, all points on the circumference are equidistant from the center. Therefore, the distance from the center O to any point on the circumference (like A, B, or C) is the radius of the circle.

4. **Conclusion about Point P**: Since PA = PB = PC, and we know that the only point that can maintain equal distances to all points on the circumference is the center of the circle, we conclude that point P must be at the center O.

5. **Finding OP**: Since both O and P are at the center of the circle, the distance OP must be equal to 0.

### Final Answer:
Therefore, OP must be equal to **0**.

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