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

Since AB is the diameter of the circle, $\angle APB={90}^{\circ }$ $(\because$ Angle in a semi circle is ${90}^{\circ })$

Thus, $△APB$ is a right-angled triangle, right angled at P. So, by Pythagoras theorem,

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

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