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Solution to Problem6


*****

Problem6:

Find coefficients a,b,c, and d so that the curve,

> ;

                           2      2
                        a x  + a y  + b x + c y + d = 0

goes through the points: (-4,5), (-2,7) and (4,-3)

Plot this curve.

SOLUTION:


Each point (x,y) produces an equation on the variables (a,b,c,d) with,

> Eq := (x,y) -> (x^2+y^2)*a + x*b + c*y + d = 0;

                                  2    2
                Eq := (x, y) -> (x  + y ) a + x b + c y + d = 0

Applying this function to the given points we get,

> Eq(-4,5); Eq(-2,7); Eq(4,-3);

                           41 a - 4 b + 5 c + d = 0

                           53 a - 2 b + 7 c + d = 0

                           25 a + 4 b - 3 c + d = 0

we can solve for a,b,c and d with,

> solve({Eq(-4,5),Eq(-2,7),Eq(4,-3)},{a,b,c,d});

                    {b = -2 a, c = -4 a, a = a, d = -29 a}

Thus, by substituting b,c and d in terms of a we obtain the curve,

> C := subs({b = -2*a, c = -4*a, d = -29*a},Eq(x,y));

                         2    2
                  C := (x  + y ) a - 2 x a - 4 a y - 29 a = 0

so for any a not 0 we can divide through by a to obtain,

> C := simplify(subs(a=1,C));

                             2    2
                       C := x  + y  - 2 x - 4 y - 29 = 0

this is the equation of a circle on the xy plane of radius R centered at the point (x0,y0)...

> Cir := expand( (x-x0)^2 + (y-y0)^2 - R^2 = 0);

                     2              2    2              2    2
             Cir := x  - 2 x x0 + x0  + y  - 2 y y0 + y0  - R  = 0

and we can see (comparing C with Cir) that, (x0,y0) = (1,2) and the radius is,

> solve(R^2-5=29,R);

                                   1/2     1/2
                                 34   , -34

since R > 0 we have that the curve is a circle on the xy plane centered at (1,2) of radius sqrt(34). Can you plot it?


Link to the commands in this file
Carlos Rodriguez <carlos@math.albany.edu>
Last modified: Wed Feb 9 09:32:29 EST 2000