Find elements of a matrix such that its determinant is zero

[Clandry]

The problem says 'Describe the set'. If I were grading the problem, the description of the set as the circle that passes through the points $a$, $b$ and $c$, along with the reasons you think so would be enough. I don't think the actual equation of the circle is needed.

Oh I see. Originally I thought the following would also satisfy the problem:
$[x_1 x_2]^T=u[a_1 a_2]+v[b_1 b_2]+w[c_1 c_2]$ for some arbitrary constants u,v,w, such that not all of them are zero.

 

[Dick]

Oh I see. Originally I thought the following would also satisfy the problem:
$[x_1 x_2]^T=u[a_1 a_2]+v[b_1 b_2]+w[c_1 c_2]$ for some arbitrary constants u,v,w, such that not all of them are zero.

I don't think that's the best way to describe the solution.

 

[Clandry]

I don't think that's the best way to describe the solution.

Yes, I thought it was too simple to describe it like that as I did not go about finding constants u,v,w.

For the equation of the circle, it seems I still am not finding the constants, but seems a bit more informative than what I had formerly planned on doing.

 

[Clandry]

No, the triangle is a given. $a$, $b$ and $c$ are ANY points.
Right. But you worked it through independently.

I just thought of something. Couldn't the coefficients in front of the $x_1^2$ and $x_2^2$ be zero? If so, then the determiannt expansion when set to zero would yield an equation for a line that passes through those 3 distinct points.

 

[Dick]

I just thought of something. Couldn't the coefficients in front of the $x_1^2$ and $x_2^2$ be zero? If so, then the determiannt expansion when set to zero would yield an equation for a line that passes through those 3 distinct points.

Yes, you might also give some thought to 'degenerate cases'. If the three points $a$, $b$ and $c$ are collinear then the 'circumcircle' becomes a line. And yes, the coefficient of $x_1^2$ and $x_2^2$ will be zero.