Points on a Circle: Find Value of 'a

[Panphobia]

Homework Statement

For what values of a do the points (4,3),(-3,1),(1,a), and (1,5) lie on a circle?

The Attempt at a Solution

So my solution came down to getting the equation a(x^2+y^2)+bx+cy+d=0
Making a matrix with the first three points, doing R1-R4, R2-R4, R3-R4, then reducing the matrix to a 3X3. After that you have the equation then sub in the last point and you get something like 0=-32a^2 - 344. My problem is that I am getting a square root of a negative number, what did I do wrong?

 

[Mark44]

Homework Statement

For what values of a do the points (4,3),(-3,1),(1,a), and (1,5) lie on a circle?

The Attempt at a Solution

So my solution came down to getting the equation a(x^2+y^2)+bx+cy+d=0
Making a matrix with the first three points, doing R1-R4, R2-R4, R3-R4, then reducing the matrix to a 3X3. After that you have the equation then sub in the last point and you get something like 0=-32a^2 - 344. My problem is that I am getting a square root of a negative number, what did I do wrong?

The only problem I see is that a is a coordinate of one of the points -- (1, a) -- and is also a variable in your equation -- a(x^2+y^2)+bx+cy+d=0. That could cause some confusion. Other than that, your overall strategy seems sound.

In row reducing a matrix, there are lots of opportunities for errors in arithmetic, so check the work you did. If you work has no errors, and you end up with an equation with no real solutions, it must be that the four points don't lie on any circle.

 

[sjb-2812]

Could you plot these out, and using the three sets of known coordinates (call these X,Y,Z) generate equations for the perpendicular bisectors of the lines XY, XZ, and YZ. Then calculate the intersection of the bisectors, which is the centre of the circle, if it exists. You also have a number of points, so you can calculate the radius; then plugs that back into th main circle equation?

 

[Panphobia]

Yea I figured it out, thanks mark, my arithmetic was wrong!