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, then 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. The angle subtended by an arc at the center is twice the angle subtended by it at any point on the remaining part of the circle. (here that point is P).

$\angle AOB={180}^{\circ }$
$\Rightarrow \angle APB=\frac{\angle AOB}{2}={90}^{\circ }$

$\therefore △APB$ is a triangle, right angled at $P$.

$\Rightarrow$ By pythagoras theorem,
${AB}^2={AP}^2+{PB}^2={\left(2PB\right)}^2+{PB}^2=5{PB}^2$

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

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