Question

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

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

Answer 1

The correct option is B

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

Given AB is a diameter of the circle.

We know that diameter subtends right angle at the circumference and hence $\angle APB={90}^{\circ }$

$⟹$ ΔAPB is a right angled triangle, right angled at P.

So, by Pythagoras theorem,

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

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

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