AB is the diameter of the circle with centre O. P is a point on the circle such that PA = 2PB. If AB = d units, find BP.

  • BP = d / √5

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The correct option is B: $\mathbf{\frac{d}{\sqrt{5}}}$

Let $O$ be the center of the circle. Since the angle subtended by the arc AB at the center is double the angle subtended in the other segment, we have:

$$ \angle AOB = 180^\circ $$


$$ \angle APB = \frac{\angle AOB}{2} = 90^\circ $$

This makes $\triangle APB$ a right-angled triangle, with the right angle at $P$. Using the Pythagorean theorem, we have:

$$ AB^2 = AP^2 + PB^2 $$

Given that $PA = 2PB$, we substitute $2PB$ for $AP$:

$$ AB^2 = (2PB)^2 + PB^2 $$

Simplifying this, we get:

$$ AB^2 = 4PB^2 + PB^2 = 5PB^2 $$


$$ AB = \sqrt{5} \cdot PB $$

Solving for $PB$:

$$ PB = \frac{AB}{\sqrt{5}} = \frac{d}{\sqrt{5}} \text{ units} $$

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