I Family of Circles through Two Points: Exists Any Circle Beyond?

[Devil Moo]

Supposed 2 circles $C_1 : x^2 + y^2 + {f_1}x + {g_1}y + h_1 = 0$ and $C_2 : x^2 + y^2 + {f_2}x + {g_2}y + h_2 = 0$ through two points.

A family of circles can be constructed as $x^2 + y^2 + {f_1}x + {g_1}y + h_1 + k(x^2 + y^2 + {f_2}x + {g_2}y + h_2) = 0$.

By altering the k, an infinitely number of circles through those two points are obtained. So is any circle existed that it cannot be obtained by the family of circles?

 

[BvU]

Well, for what value of $k$ does one get back $C_2$ ?

 

[Devil Moo]

Well, for what value of $k$ does one get back $C_2$ ?

Let $C = \{C_1, C_2, C_3, ...\}$ be the set of equations of circles through those two points.
Can we get all the elements of $C$ from the family of circles?

 

[BvU]

Well, for what value of $k$ does one get back $C_2$ ?

You get back $C_1$ when choosing $k=0$. But $C_2$ ?

 

[Devil Moo]

You get back $C_1$ when choosing $k=0$. But $C_2$ ?

Um... choosing $\lim_{k \rightarrow \infty} k$?

 

[mfb]

You cannot do that, k has to be a real number. Do the variables have to have the same value for C2 and the class of circles?

If all the variables apart from k are fixed in the family of circles, then there are certainly circles not included in that family - your family has just one degree of freedom while the set of all circles in a plane has three.

 

[Devil Moo]

$f_1, f_2, g_1, g_2, h_1, h_2$ are fixed.

Do $x, y$ count as degree of freedom?Do $f, g, h$ have relationship to make circle passing through those points?

 

[BvU]

No, $x$ and $y$ are subject to the equations. $f_1, f_2, g_1, g_2, h_1, h_2$ are six degrees of freedom, but four of those are needed to let the circles go through the two points $(x_1,y_1)$ and also through $(x_2, y_2)$. The other two can be found from the radii of $C_1$ and $C_2$.

In the family of circles there is one degree of freedom (k). Makes sense, because the centers of the circles are constrained to be located on the perpendicular bisector of the line segment from $(x_1,y_1)$ to $(x_2, y_2)$.

But $C2$ isn't in the family...

 

[Devil Moo]

But $C2$ isn't in the family...

How about modify the argument that exclude $C_2$?

 

[BvU]

I am not sure what you mean with that question ?

 

[mfb]

Do x, y count as degree of freedom?

No.

Your family of circles is not necessarily a family of circles, by the way. Many pairs of circles lead to empty sets for the second equation.

 

[Devil Moo]

I am not sure what you mean with that question ?

How about now $C = \{ C_1, C_3, C_4, ...\}$?
Can we get all the elements of $C$ from the family of circles?

Your family of circles is not necessarily a family of circles, by the way. Many pairs of circles lead to empty sets for the second equation.

Sorry. I do not get it.

 

[mfb]

Sorry. I do not get it.

As example, consider:
C1: $x^2+y^2-4x-4y+7=0$
C2: $x^2+y^2+4x+4y+7=0$
For k=1, how does the corresponding figure look like?
C_k=1: $2x^2+2y^2+14=0$. Try to find solutions for that.

 

[BvU]

Not sure I like this example: C₁ and C₂ have no points in common...

I'm embarking on some real work now (yuck, against my nature) by bringing in another example:

$C_1 :\ x^2 + (y-1)^2 = 5\quad$ and $\ C_2 :\ x^2 + (y+1)^2 = 5\quad$ with $\ (-2,0)\$ and $\ (2,0) \$ in common.

For $k = -1$ we get no solution, for $k\ne -1$ we get
$$C_k:\ x^2 + \left (y- {1-k\over 1+k}\right )^2 = 5 + \left ({1-k\over 1+k}\right )^2$$ if I'm not mistaken.
Circles through the two common points with center $\ \left (0,{1-k\over 1+k} \right )\$.

That means we found an infinity of circles trough the two common points, but not all of them:
we don't have $C_2$ in the family, and got one black sheep (a no solution non-circle) in return.

 

[mfb]

Ah, the "through two points" mean they intersect. Okay, then it works, k=-1 as exception (where the equation becomes linear - it has solutions, but not a circle).

 

[BvU]

it has solutions

You're right - I made a sign error and overlooked that $y=\pm {1\over 2}$ solves $C_{-1}$. Funny case.

 

[disregardthat]

The family of circles going through the two points in $C_1 \cap C_2$ is given by $k_1(x^2+y^2 + f_1x+g_1y+h_1) + k_2(x^2+y^2 + f_2x+g_2y+h_2) = 0$, where $k_1$ and $k_2$ are not both zero. You retrieve the same circle whenever $k_1$ and $k_2$ are scaled by some constant $\lambda \not = 0$. In other words, such a circle uniquely determines a point $(k_1:k_2)$ on the projective line $\mathbb{P}^1$ and vice versa. The degenerate case is the point $(k_1:k_2) = (1:-1)$, in which case you end up with a line. In this sense, the affine line $\mathbb{A}^1 \cong \mathbb{P}^1 \backslash \{(1:-1)\}$ parametrizes the family of circles going through these two points.

 

[Devil Moo]

Circles through the two common points with center $\ \left (0,{1-k\over 1+k} \right )\$.

When I construct a equation of circle passing through $(-2, 0)$ and $(2, 0)$ which is not $C_2$, that equation can be retrieved by $C_k$. If we plot the center of all circles passing through two points, I think it must be a straight line.

The family of circles going through the two points in $C_1 \cap C_2$ is given by $k_1(x^2+y^2 + f_1x+g_1y+h_1) + k_2(x^2+y^2 + f_2x+g_2y+h_2) = 0$, where $k_1$ and $k_2$ are not both zero.

By this expression, does it obtain all circles and straight line passing through 2 points?

 

[BvU]

You're right - I made a sign error and overlooked that $y=\pm {1\over 2}$ solves $C_{-1}$. Funny case.

How come no one puts me right there ? $(y+1)^2 - (y-1)^2 = 0\quad$ gives y = 0. One single straight line through both points in the example. No 'funny case' at all, just a sensible limiting case.

If we plot the center of all circles passing through two points, I think it must be a straight line.

Yep, the y-axis in the example. Minus the point $(-1,0)$

 

[disregardthat]

When I construct a equation of circle passing through $(-2, 0)$ and $(2, 0)$ which is not $C_2$, that equation can be retrieved by $C_k$. If we plot the center of all circles passing through two points, I think it must be a straight line.By this expression, does it obtain all circles and straight line passing through 2 points?

Yes. The degenerate case can also be thought of as a generalized circle.

 

[Devil Moo]

Is k served as a parameter to choose the center points of the circle passing through those two points, for k != -1?

 

[BvU]

Yes.