Thread: Find equation of a circle given three points without using determinants

I am solving the problems of an elementary linear algebra textbook. In the section where Gaussian elimination is first introduced as a method to solve systems of equations with simple row operations there is this question:

Find coefficients a, b, c and d so that the curve shown in the accompanying figure is given by the equation ax^2+ay^2+bx+cx+d=0. Then there is a graph of the circle with three points marked: (-4,5), (-2,7) and (4,-3).

This is how I solved it:

First I used the coordinates of the three points to write down a linear system with four unknowns and three equations. I then wrote the matrix in reduced row echelon form and expressed each of the coefficients in terms of d. I noticed that using the equation of the circle I could express x in terms of y, so I did this and substituted the values of x and y for the points (-4,5) and (4,3) and expressed the constants in terms of d. Next I multiplied the expression for -4 by -1 and thus got two expressions for 4 in terms of d. Finally I added both expressions and simplified for d.

In that way I found one solution: a=-1, b=2, c=4, d=29. I think everything is fine except that I did not know how to prove that the sign of the root involved in the expression of -4 (in terms of y) was negative and that it was positive for the expression of 4. I only guessed this form the diagram. Could someone please tell me how to do this?

I found that there is another method to solve this problem, which is called by determinants, and that gives the equation as a result. However, we have not covered that in class yet, and this problem is at the very beginning, even before matrix algebra, so it seems that it is expected that one solves the problem only using row operations. This was the simplest way I could find to solve the problem by using row operations without finding the centre of the circle, which would be another method which, I think, is not what is intended for this exercise.

Do my question is, did I do something really stupid and waste my time? Can somebody else think of a simpler solution with the restrictions that I have listed?

Thank you very much!

Have you considered constructing perpendicular bisectors of any two chords?

Given that the figure is a circle,
can you not simply use a=1 ?
as the Cartesian equation of a circle is (x-h)^2+(y-k)^2=r^2
which when expanded, gives your given equation with a=1.

then you have 3 sets of (x,y) values to substitute
to find the remaining 3 constants b, c and d.

As things stand, you have 3 sets of values to solve for 4 unknowns
and you can use 3 clues to find 3 unknowns,
so there is certainly another clue you need to use.

Originally Posted by Archie Meade

Given that the figure is a circle,
can you not simply use a=1 ?
as the Cartesian equation of a circle is (x-h)^2+(y-k)^2=r^2
which when expanded, gives your given equation with a=1.

then you have 3 sets of (x,y) values to substitute
to find the remaining 3 constants b, c and d.

As things stand, you have 3 sets of values to solve for 4 unknowns
and you can use 3 clues to find 3 unknowns,
so there is certainly another clue you need to use.

Well, thank you, this seems to be a much simpler way of doing it. The problem is a bit deceptive, though, you need to remember the fact that the coefficient of the squares in the equation of a circle is one, so actually you are only looking for three variables. It seems that I need to review my analytic geometry! What about what I did, do you know of a way of proving the signs of the root in the equation without finding first the centre of the circle?