Solution 3 (a).

See text and/or instructor's solution manual.

Answer.   [Graphics:../Images/JoukowskiTransModHome_gr_103.gif].  

Solution.   The formula for a circle with center   [Graphics:../Images/JoukowskiTransModHome_gr_104.gif]   is

                    [Graphics:../Images/JoukowskiTransModHome_gr_105.gif].  

Using the point   [Graphics:../Images/JoukowskiTransModHome_gr_106.gif]   we have   [Graphics:../Images/JoukowskiTransModHome_gr_107.gif].  

Therefore,  

                    [Graphics:../Images/JoukowskiTransModHome_gr_108.gif].  

 

We are done.   

 

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/JoukowskiTransModHome_gr_109.gif]

[Graphics:../Images/JoukowskiTransModHome_gr_110.gif]


[Graphics:../Images/JoukowskiTransModHome_gr_111.gif]

[Graphics:../Images/JoukowskiTransModHome_gr_112.gif]


[Graphics:../Images/JoukowskiTransModHome_gr_113.gif]

[Graphics:../Images/JoukowskiTransModHome_gr_114.gif]


We are really done.   

 

Aside.  We can graph some of the circles.  

                    [Graphics:../Images/JoukowskiTransModHome_gr_115.gif]          [Graphics:../Images/JoukowskiTransModHome_gr_116.gif]

  

                    [Graphics:../Images/JoukowskiTransModHome_gr_117.gif]          [Graphics:../Images/JoukowskiTransModHome_gr_118.gif]

                    The circles   [Graphics:../Images/JoukowskiTransModHome_gr_119.gif]   for   [Graphics:../Images/JoukowskiTransModHome_gr_120.gif]   are    

                    [Graphics:../Images/JoukowskiTransModHome_gr_121.gif].

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) 2008 John H. Mathews, Russell W. Howell