Krdytkyu

In learning about the general formula for a circle:
$$(x-h)^2 + (y-k)^2 = r^2$$
my book states that $3$ points are sufficient to guarantee the solution (or absence of solution) for an unique circle.

It follows this up by explaining that $3$ different points creates $3$ different equations...namely:

1) $(x_1-h)^2 + (y_1-k)^2 = r^2;$ where $;(x_1,y_1);$ is the first point provided

2) $(x_2-h)^2 + (y_2-k)^2 = r^2;$ where $;(x_2,y_2);$ is the second point provided

3) $(x_3-h)^2 + (y_3-k)^2 = r^2;$ where $;(x_3,y_3);$ is the third point provided

I'm okay with this as I'm familiar with the whole "$3$ equations for $3$ unknowns".

However, the book's explanation concludes with the statement:

"The equations can't be linearly dependent since the points that lead to them are all different". I am not particularly familiar with linear algebra so I do not understand the significance of this statement. Is there a visual representation of this statement that can help me understand what they mean?

Sometimes a few figures are worth a thousand words:

The pictures in @DavidGStork 's lovely answer explain why the three equations produce a unique solution.

I want to add a discussion of the algebra, and a possible correction.

When the author writes

The equations can't be linearly dependent since the points that lead
to them are all different

the intent is to tell you that you can't deduce the third equation from the first two. That means you really do have three independent conditions on the three unknowns, which is usually what you need to guarantee a solution.

However

• The simple assertion that the "points are different" does not in
  itself prove the independence. For that you need an argument like the
  one in the picture.




• As written, these are not linear equations. The variables $h$ and $k$ appear
  squared. So they should be described as independent, but not linearly
  independent. With a little algebra you can combine them in pairs to produce two linear equations for $h$ and $k$. Those you can solve using what you learn in high school algebra. Finding where two lines cross by solving two equations in two unknowns is the only "linear algebra" you need. Then you can use any one of the three equations to find $r^2$, so $r$.

This doesn’t have to do with linear algebra per se, although the idea of linear independence is central to that discipline. To state the claim differently, none of the three equations is redundant—none of them can be obtained by adding multiples of the other two to each other—because the three points are distinct.

Assume that the equations are dependent. W.l.o.g. we’ll say that it’s the third one that’s a linear combination of the other two. Since its right-hand side must be $r^2$, there’s some real number $lambda$ for which the coefficients of the left-hand side of $$(1-lambda)((x_1-h)^2+(y_1-k)^2)+lambda((x_2-h)^2+(y_2-k)^2) = r^2 tag{*}$$ are equal to those in the left-hand side of the original third equation. Now, that equation represents a circle with radius $r$ centered at $(x_3,y_3)$. With a bit of work, equation (*) can be rearranged into $$(h-((1-lambda)x_1+lambda x_2))^2 + (k-((1-lambda)y_1+lambda y_2))^2=r^2-lambda(1-lambda)((x_1-x_2)^2+(y_1-y_2)^2).$$ This is an equation of a circle with center at $(1-lambda)(x_1,y_1)+lambda(x_2,y_2)$, which lies somewhere on the line through $(x_1,y_1)$ and $(x_2,y_2)$. For its radius to be $r$, the second term on the right must vanish. This can occur in three ways: either $(x_1,y_1)=(x_2,y_2)$; $lambda = 0$, in which case $(x_3,y_3)=(x_1,y_1)$; or $lambda=1$, in which case $(x_3,y_3)=(x_2,y_2)$. By hypothesis, the three points are distinct, so the three equations must be linearly independent.