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

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- #1

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

- #2

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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].

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HallsofIvy

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HallsofIvy

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- #6

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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

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mathwonk

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