Chapter 10, Problem 14RP

To determine

To find : Length of common external tangent AB.

Answer to Problem 14RP

  $8$

Explanation of Solution

Given information :

As shown in the figure OA = $8$ , PB = $2$ and construct PQ s.t. PQ = AB.

  

Formula Used:

Angle between tangent and radius = $90°$

Calculation :

OD = $8$ (Radius)

PD = $2$

  $\therefore$ OP = $8+2=10$

In $\Delta \text{OPQ,}$

By Pythagoras Theorem,

  $\begin{array}{l}O{P}^2=O{Q}^2+P{Q}^2\\ \Rightarrow PQ=\sqrt{{10}^2-6^2}=8\\ \therefore AB=PQ=8\end{array}$