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If you have two circles that always have equal radius on a plane, and the circles have different center points, both on the x axis to make it simpler. Such that they intersect twice.

As the radius increases the distance between the points of intersection will increase, and they will be almost opposite each other on the circle, but never quite reach that point.

As the radius tends towards infinity the distance between the edges if the circles will get smaller and smaller, so it will tend towards zero, and so the circles will overlap and must be the same.

The reason I ask is because: where z is a complex number, $|z - a| = |z - b|$, where $a \neq b$ and a and b are real numbers (to make it simpler as above), is a locus, the perpendicular bisector of the line ab. (This is because it can be thought of as all the points where circles around the points a and b (each) intersect, in the same way you'd draw a perpendicular bisector with a compass)

So in addition to the perpendicular bisector, won't the locus also include a circle of infinite radius?

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Although the angle formed at the intersection of the two circles will tend to zero as their radii tends to infinity, the circles will not overlap. Given $a$ and $b$ have a distance of $|a-b|$, the distance between the edges of each circle along the line formed between $a$ and $b$ will always be $|a-b|$, no matter what the radius is, given the radii of both circles are always the same.

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  • $\begingroup$ But there is a similar thing with differentiation, where the width of each "slit" tends to zero. And in the same way 0.9 recurring equals 1? $\endgroup$
    – Jonathan.
    Mar 5, 2013 at 11:50
  • $\begingroup$ I get where you're coming from in saying this, but the concepts are different... the idea of a limit, in general, is that an expression approaches, but never reaches, a value. This is different, because the distance between the edges never even approaches 0 - it remains a constant |a-b|. As @politopo said, the distance will appear to become smaller, because the circles are getting larger, but the absolute value of the distance will not change at any point $\endgroup$
    – mardat
    Mar 5, 2013 at 12:27
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The asnwer mardat gave is correct, nevertheless, when you look at the circles from far above, i.e. if you consider the distance $|a-b|$ normalized with the radius $r \rightarrow \infty$, the normalized distace $\frac{|a-b|}{r}$ tends to zero. From this viewpoint, somehow the two circles tendo to overlap.

The fact is, Jonhatan, that when you look at a circle which has infinite radius and is centered at infinity, you see a straight line, which happens to be the bisector you already have.

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  • $\begingroup$ Can you expand on your last paragraph? $\endgroup$
    – Jonathan.
    Mar 5, 2013 at 11:51
  • $\begingroup$ A circle whose center tends to infinity and which has a point in a finite position, tends to a straight line. This is easy to see if you look at the equation of a circumference $(x-c)^2 + y^2 = c^2$, or $(x-c)^2/c^2 + y^2/c^2 = 1$ for $\lim x\rightarrow \infty$, it tends to become $(x-c)^2/c^2 = 1$, that is the (squared) equation of a straight line. $\endgroup$
    – politopo
    Mar 6, 2013 at 10:43

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