Someone explain continuity principle...

scout6686

http://chutzpah.typepad.com/.a/6a00e...75cf644970b-pi

How do the circles still intersect at the bottom, and at 2 points like the top 2 circles?

micromass

You can show this algebraically. Let's take our circle with radius 1.
Then the red circle has center at (0,0) and has radius 1. The equation for such a circle is
[tex]x^2+y^2=1[/tex]
The blue circle has center at (5,0) and has radius 1. The equation is
[tex](x-5)^2+y^2=1[/tex]

We can now find the points in the intersection of these two circles. We know from the first equation that

[tex]y^2=1-x^2[/tex]

Substituting that in the second equation gets us

[tex](x-5)^2 + (1 -x^2 )=1[/tex]

This is an equation that can easily be solved. we get x=5/2. We substitute that in the first equation and get
[tex]y^2=-21/4[/tex]

and thus

[tex]y=\pm i\sqrt{21}/2[/tex]

So the points of intersection are [itex](5/2,i\sqrt{21}/2)[/itex] and [itex](5/2,-i\sqrt{21}/2)[/itex].

HallsofIvy

But the y values are imaginary numbers while the numbers defining the coordinate system must be real numbers- so to say the circles "intersect" there is generalizing "intersect" a heck of a lot!

Monsterboy

is it possible to plot the circles with y-axis having the imaginary part and x axis having the real part(on the complex plane)?

HallsofIvy

[itex]e^{R\theta}[/itex] gives a circle with center at 0 and radius R in the complex plane. You cannot plot an equation like y= f(x) with y and x complex numbers because you would have to have real and complex axes for both x and y- and that requires 4 dimensions.

tensor33

It is my understanding that the intersection does exist, just not in the euclidian plane. So it can be said that the circles intersect without changing the meaning of intersection

mathwonk

this principle in its simplest form says that the equations X^2 = t always have two solutions no matter what t is. if you believe that, then you must also believe the original assertion, as micromass showed.