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. √2d
• B. d/√5
• C. √5d
• D. 2√2d

Answer 1

The correct option is B

d/√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 and
∠AOB = ${180}^{\circ }$, $\angle APB=\frac{1}{2}\angle AOB={90}^{\circ }$,ΔAPB is a right angled triangle, right angled at P.
So, By Pythagoras theorem,${AB}^2$ = ${AP}^2$ + PB²
= ${\left(2PB\right)}^2$ + ${PB}^2$
= 5 ${PB}^2$
$\Rightarrow$ AB = $\sqrt{5}$PB$\Rightarrow$ PB = $\frac{AB}{\sqrt{5}}$ = $\frac{d}{\sqrt{5}}$