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.
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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? |