Question

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

• A. $\sqrt{2}d$
• B. $\frac{d}{\sqrt{5}}$
• C. $\sqrt{5}d$
• D. $2\sqrt{2}d$

Answer 1

The correct option is B

$\frac{d}{\sqrt{5}}$

Let O be the centre of the circle. Since angle subtended by the arc AB at the centre is half the angle subtended in the other segment.

∠AOB = ${180}^{\circ }$

∠APB = $\frac{1}{2}$ ∠AOB = ${90}^{\circ }$

ΔAPB is a right triangle, right angled at P. So, by Pythagoras theorem,

${AB}^2={AP}^2+{PB}^2$
= ${\left(2PB\right)}^2+{PB}^2$
= $5{PB}^2$

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

$\Rightarrow$ $PB=\frac{AB}{\sqrt{5}}=\frac{d}{\sqrt{5}}$