Chapter 10.4, Problem 14PSB

To determine

To find : The length of a common internal tangent.

Answer to Problem 14PSB

  $\text{Length of a common internal tangent = 9}\text{.75 units}$

Explanation of Solution

Given information : $\text{The centres of two circles with radii 3 and 5 are 10 units apart}\text{.}$

  

  $\begin{array}{l}\text{Let }{C}_1\text{ and }{C}_2\text{ are the centres of radii 5 and 3 and P is intersection}\\ \text{of }{C}_{\text{1}}{C}_2\text{ and a common internal tangent to circles}\text{.}\\ \text{In }\Delta C_{1}AP\text{and}{C}_{2}BP\\ \text{Let }{C}_{1}P=a\text{ and }P{C}_2=10-a\\ \angle C_{1}AP=\angle C_{2}BP\text{(90°)}\\ \angle {\text{APC}}_{\text{1}}=\angle BP{C}_2\text{(Vertically opposite)}\\ \therefore \Delta C_{1}AP~\Delta C_{2}BP\text{(AA similarity)}\\ \text{Hence,}\\ \frac{\text{3}}{\text{1}0-a}=\frac{5}{a}\Rightarrow 3a=5(10-a)\\ \Rightarrow 3a=50-5a,\text{So }a=6.25\\ \therefore C_{1}P=6.25\text{ and }P{C}_2=3.75\text{ units}\\ AP=7.5\text{ units and }BP=2.25\\ \Rightarrow AB=9.75\text{ units}\end{array}$