Someone explain continuity principle

[scout6686]

http://chutzpah.typepad.com/.a/6a00e55180ed5c88340120a75cf644970b-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
$$x^2+y^2=1$$
The blue circle has center at (5,0) and has radius 1. The equation is
$$(x-5)^2+y^2=1$$

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

$$y^2=1-x^2$$

Substituting that in the second equation gets us

$$(x-5)^2 + (1 -x^2 )=1$$

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

and thus

$$y=\pm i\sqrt{21}/2$$

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

 

[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]

$e^{R\theta}$ 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]

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!

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.