# Someone explain continuity principle

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
Homework Helper
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!

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
Homework Helper
$e^{R\theta}$ 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.

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