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I am reading a paper, and in it it says "disjoint circles of a coaxial system". What does the writer mean by that?

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I'm pretty sure this is the most recent excuse Comcast gave me as to why my internet wasn't working at the advertised speeds... – Emily Mar 27 '13 at 17:30
up vote 2 down vote accepted

Coaxial means it shares an axis. You can imagine these as circles that have either the same centre in 2D or whose centres belong to a common line in 3D (where the circles are probably assumed to be parallel if so, check that).

Edit: As Lord Farin has said, it could also very well mean that the centres all lie on the same line even in $2D$. Go through the paper a bit more and see if you can figure it out.

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We're talking about a 2D world here. I assumed it meant circles that have the same center; why the word 'disjoint' then? – George Mar 27 '13 at 17:30
Probably to eliminate the trivial case of equal radii – muzzlator Mar 27 '13 at 17:31
That's what I thought. Thanks! – George Mar 27 '13 at 17:32
It should not be without consideration that the circles may not be concentric, but rather have e.g. the $x$-coordinate in common. – Lord_Farin Mar 27 '13 at 17:33
This actually sounds more plausible than my answer, thanks. I'll update it. @Lord_Farin – muzzlator Mar 27 '13 at 17:34

A coaxial system of circles is a set of circles in a plane, any two of which share a common radical axis: a line which is the locus of points from which the tangents to any two of the circles are equal in length. In the case that two of the circles intersect, they all do, and the radical axis is the line through the common pair of intersection points for all the circles. It is true, as others have said, that the centres of the circles of such a system lie on a common line; but this condition is not sufficient to define a coaxial system of circles.

Example: Fix a nonzero real number $\lambda$. Then the circles $$x^2-2ax+y^2=\lambda\quad (a\in \Bbb R; a\neq 0)$$comprise a coaxial system of circles with radical axis $x=0$. If $\lambda > 0$, then they all meet at $(0, \pm\sqrt \lambda )$.

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