Chapter 10.1, Problem 23PSC

To determine

$PS(\text{the distance from P to }\odot O)$

Answer to Problem 23PSC

  $2$

Explanation of Solution

Given information : $\begin{array}{l}PQ=4,RQ=10,PO=15\text{ circle centre O}\text{.}\\ \text{Draw OT, OT }\perp \text{ RQ}\text{. Draw OQ}\text{.}\end{array}$

  

  $\begin{array}{l}RQ=10\Rightarrow RT=TQ=5\\ \text{In triangle OTP,}\\ {\text{PO}}^{\text{2}}=T{P}^2+O{T}^2\\ {15}^2=9^2+O{T}^2\\ 225-81=O{T}^2\\ \Rightarrow O{T}^2=144\Rightarrow OT=12\\ \text{In triangle OTQ,}\\ O{Q}^2=O{T}^2+T{Q}^2\\ O{Q}^2={12}^2+5^2=144+25=13\\ OQ=OS=\text{Radius}=13\\ \Rightarrow PS=15-13=2\\ PS=2\end{array}$