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O is the centre of a circle. There is a ...

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 :

A

`(PA + PB + PC)/(3)`

B

`(PA + PB + PC)/(2)`

C

`(AB + BC)/(2)`

D

zero

Text Solution

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The correct Answer is:
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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