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Let us consider two circles in the (real) plane:

$C_1 : (x-x_1)^2 + (y-y_1)^2 - r_1^2 = 0$

$C_2 : (x-x_2)^2 + (y-y_2)^2 - r_2^2 = 0$

In order to calculate their intersection point we can easily find the line defined by the two points by subtracting $C_1$ from $C_2$:

$L: -2(x_1-x_2)x + (x_1^2-x_2^2) - 2(y_1-y_2)y + (y_1^2-y_2^2) - (r_1^2-r_2^2) = 0$

Plugging this back into either circle equation $C_i$ we can easily determine the intersection points of $C_1$ and $C_2$. So we can view the line $L$ as the line defined by the two intersection points.

But is there also a nice geometric interpretation of $L$ if the circles do not intersect?

(That is, when the distance between their centers is smaller than $\min \{r_1,r_2\}$ or larger than $r_1+r_2$.)

enter image description here

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    $\begingroup$ @Downvoter: I've tried to make it as clear as possible. Could you explain how I can improve the question? $\endgroup$ – flawr Jun 10 '18 at 15:55
  • $\begingroup$ Excellent visualization!! Care to tell us how is was generated? $\endgroup$ – JonathanZ Jun 10 '18 at 16:03
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    $\begingroup$ @JonathanZ Thanks! I used Geogebra. For the center of the moving circle I created a slider with the desired range of coordinates, and you can easily animate it with the command in the popup menu when you right-click it. Then I just recorded it with LiceCap. $\endgroup$ – flawr Jun 10 '18 at 16:06
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    $\begingroup$ It is called the radical axis (or power line) of the two circles. For the potion of radical axis outside the circles, it is the locus of points at which tangents draw to both circles have the same length. $\endgroup$ – achille hui Jun 10 '18 at 16:58
  • $\begingroup$ @achillehui Please consider adding your comment as an answer, as it is a perfectly valid answer to my questions! $\endgroup$ – flawr Jun 10 '18 at 17:25
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if $C_1$ doesn't intersect $C_2$, then there is a pair of points $(A,B)$ such that $A$ is the image of $B$ by both inversions through $C_1$ and $C_2$.

Then the line you are seeing is a line going through the middle of those two points. Also, because the line joining the two centers is an axis of symmetry of the picture and since your line is defined purely geometrically, it has to stay an axis of symmetry, so your line has to be perpendicular to the axis (it can't be the axis itself or else it would have real intersection points with the circles).

There is a way to interpret $A$ and $B$ as the pair of conjugate complex points that are at the intersection of $C_1$ and $C_2$, (and so also on the line), I give a lot more info about that in this answer What is the reflection across a parabola?

That line is also the locus of points that have the same power with respect to both circles, see Line intersecting complex points of two circles

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Pass from $\mathbb{R}^2$ to $\mathbb{C}^2$, and the equations will always have solutions. In the case where the circles don't intersect, the points predicted by the equations will be imaginary.

Yet, the line between those imaginary points does intersect the real plane. That is the line you see.

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  • $\begingroup$ I am aware of the extention to $\mathbb C$, but I'm looking for a way to explain it in the real plane only. It certainly looks like $L$ is always perpendicular to the line connecting the centers of the circles, but maybe it has other nice properties? $\endgroup$ – flawr Jun 10 '18 at 16:05

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