Example 4.  Use the determinant method to find the circle through the points (6,1),  (2,2)  and  (1,4).
Remark. In Exercises 5 and 6 the same points are used to find the standard parabola and alternate parabola.

Solution 4.

The points are entered into Mathematica with the command:

 

[Graphics:../Images/ConicFitMod_gr_34.gif]

Then a row vector corresponding to equation (3) is defined:

 

[Graphics:../Images/ConicFitMod_gr_35.gif]

The matrix A for the linear system in (4) and the determinant is now created.  The vector R is stored in the first row by issuing the command  A = {R}.  Then the remaining three rows of A are generated with the loop command:

[Graphics:../Images/ConicFitMod_gr_36.gif]


[Graphics:../Images/ConicFitMod_gr_37.gif]

For the given three points, the homogeneous system AC = 0 is:

 

[Graphics:../Images/ConicFitMod_gr_38.gif]

 

The determinant of this matrix is computed by typing:

[Graphics:../Images/ConicFitMod_gr_39.gif]


[Graphics:../Images/ConicFitMod_gr_40.gif]

The desired equation is:

[Graphics:../Images/ConicFitMod_gr_41.gif]


[Graphics:../Images/ConicFitMod_gr_42.gif]

The conic is the circle shown in Figure 4.  It is plotted using the commands:

[Graphics:../Images/ConicFitMod_gr_43.gif]


[Graphics:../Images/ConicFitMod_gr_44.gif]

[Graphics:../Images/ConicFitMod_gr_45.gif]

[Graphics:../Images/ConicFitMod_gr_46.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) John H. Mathews 2004