Math Is Fun Forum

Dragonshade

Member

Registered: 2008-01-16

Posts: 147

Intersections

how to prove that an arbitrary circle only intersects y=1/x (x>0) at most 2 points. I tried brute force, polynomial equation with degree 4 is not fun at all. Curvature might be a way, but I have no ideas : (

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Intersections

Hi

This is how I would do it.

substitute 1 / x into x^2 + y^2 = r^2.

You will get a quartic equation that can be converted to quadratic and solved for.

Four roots 2 positive and 2 negative. Which is correct. We expect 4 intersections if you do not have the constraint x > 0.

Since x>0 you can eliminate the two negative roots.

So there are only 2 points of intersection.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Dragonshade

Member

Registered: 2008-01-16

Posts: 147

Re: Intersections

I got the equation for intersection is  (x-a)^2 +( 1/x -b)^2= r^2 , for circle centered at (a,b). And its of degree 4th. So it should have 4 roots. so At most 4 real roots. but the tricky part is that, y=1/x has two symmetric parts. x>0 and x<0. So the 4 intersections should be 2 for each one. I wish I just could argue from the symmetry but I dont know the mathematics for that.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Intersections

Hi Dragonshade;

Please look at my post. I believe that problem has been solved. At least for the arbitrary circle at (0,0).

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Dragonshade

Member

Registered: 2008-01-16

Posts: 147

Re: Intersections

Ah, sorry, I didnt see that cuz the page didnt load fully. I am looking at it

But I mean any arbitrary circle, not necessary the one center at the origin.

Last edited by Dragonshade (2011-01-22 17:04:58)

gAr

Member

Registered: 2011-01-09

Posts: 3,482

Re: Intersections

Hi Dragonshade,

You can check from the graph.
Since x>0, the symmetrical part for negativeve x is eliminated.

---

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense"  - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Intersections

Yes, but mine only covers a circle centered at the origin.

Visually you can see it but that is not quite a proof.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Dragonshade

Member

Registered: 2008-01-16

Posts: 147

Re: Intersections

a full fledged quartic is just nightmare.

is there any theorem about the intersection of curves based on curvature

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Intersections

Hi;

Not that I can remember offhand. I will post if I find a general solution.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.