Question

In this figure, the centre of the circle is $O$. $AB\perp BC$, $ADOE$ is a straight line, $\overline{AP}=\overline{AD}$ and $AB$ has a length twice the radius. Then:

• A. $\overline{A{P}^2}=\overline{PB}.\overline{AB}$
• B. $\overline{AP}.\overline{DO}=\overline{PB}.\overline{AD}$
• C. $\overline{A{B}^2}=\overline{AD}.\overline{DE}$
• D. $\overline{AB}.\overline{AD}=\overline{OB}.\overline{AO}$
• E. None of these

Answer 1

The correct option is A $\overline{A{P}^2}=\overline{PB}.\overline{AB}$

Since AB is tangent to the circle, we have $\overline{AD}|\overline{AB}=\overline{AB}|\overline{AE},$
$\overline{A{B}^2}=\overline{AD}.\overline{AE}$. But $\overline{AE}=\overline{AD}+2r=\overline{AD}+\overline{AB};$
$\therefore \overline{A{B}^2}=\overline{AD}(\overline{AD}+\overline{AB})=\overline{A{D}^2}+\overline{AD}.\overline{AB}$
$\therefore \overline{A{D}^2}=\overline{A{B}^2}-\overline{AD}.\overline{AB}=\overline{AB}(\overline{AB}-\overline{AP})=\overline{AB}.\overline{PB}.$
Since $\overline{AP}=\overline{AD},\overline{A{P}^2}=\overline{AB}(\overline{AB}-\overline{AP})=\overline{AB}.\overline{PB}.$