Elementary Linear Algebra: Finding the coefficients to a,b,c, and d of a given curve.

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"Find the coefficients a,b,c, and d so that the curve shown in the accompanying figure [a circle in the xy-plane that passes through the points (-4,5),(-2,7),(4,-3)] is given by the equation ax^{2}+ay^{2}+bx+cy+d=0"

What I've done so far is that I have used the given points and substituted them into the given equation thus giving me three equations and four unknowns:

- 41a-4b+5c+d=0
- 29a-2b+7c+d=0
- 25a+4b-3c+d=0

In which have then three equations into an augmented matrix and proceeded into putting it into a reduced row echelon form. While doing so, I realized that the numbers where quite off and that I was no further into the solution to the problem.

Please, if anybody, pinpoint where I went wrong and guide me toward the solution and why it is the solution.

Thanks so much.

Re: Elementary Linear Algebra: Finding the coefficients to a,b,c, and d of a given cu

Since the graph of the equation is a circle a cannot be 0. So we will divide the equation by a to get the equation as

x^2 + y^2 + (b/a) x + (c/a) y + d/a = 0 For easy handling we may put b/a = P, c/a = Q and d/a = R. Thus the equation becomes

x^2 + y^2 + P x + Q y + R = 0 , Now we have three variables and three points and hence we can solve it.