- #1

Gerenuk

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After playing around I had various equations. For example the problem is equivalent to finding beta in

[tex]\Re\left((\exp(i\beta)+z)^2\right)=b[/tex]

where z is complex. Suggestions?

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- Thread starter Gerenuk
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- #1

Gerenuk

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After playing around I had various equations. For example the problem is equivalent to finding beta in

[tex]\Re\left((\exp(i\beta)+z)^2\right)=b[/tex]

where z is complex. Suggestions?

- #2

tiny-tim

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I don't know whether this will work, but have you tried simplifying it by squashing one of the coordinate axes so that one of the ellipses is a circle?

- #3

Gerenuk

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The above equation looks the simplest I could find. Yet not sure how to proceed.

- #4

fzero

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With two ellipses, it seems like we can simplify things quite a bit. Use linear transformations to map one of the ellipses to a

For the ellipses to only touch, these four points must be coincident. This can only be the case if [tex]r_+=r_- = \pm 1[/tex]. This requires the discriminant of the quadratic to vanish as well as a condition on the ratio of coefficients.

There's a certain amount of work to do to compute the coefficients of the quadratic above in terms of the parameters of the original ellipsi, but it doesn't seem too bad.

- #5

Gerenuk

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Hmm? How does that work? The ellipse is rotated so the equation still looks complicating?!Now since the unit circle is invariant under rotations in the plane, we can use a rotation to set, say, [tex]y_0'=0[/tex]. But then the quartic equation for the intersection points is actually a perfect square so the problem reduces to solving a quadratic equation.

- #6

fzero

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Hmm? How does that work? The ellipse is rotated so the equation still looks complicating?!

After transforming the first the ellipse into a unit circle ([tex]x=\cost t, y=\sin t[/tex]), the equation for the points of intersection is

[tex] \frac{(\cos t -x_0')^2}{(a')^2} + \frac{(\sin t-y_0')^2}{(b')^2} = 1.[/tex]

After a rotation with [tex]\tan\theta = y_0'/x_0'[/tex], this is put into the form

[tex] \frac{(\cos(t-\theta)-x_0'')^2}{(a')^2} + \frac{\sin^2(t-\theta)}{(b')^2} = 1,~~(*)[/tex]

where

[tex]x_0''=x_0'\sqrt{ 1+ \left(\frac{y_0'}{x_0'}\right)^2}.[/tex]

The whole point is that (*) is a quadratic equation for [tex]cos(t-\theta)[/tex].

- #7

Gerenuk

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To clarify, I tried plugging in t=a+b and finding an equation for b such that only cos(a) and cos^2(a) terms remain. However, this is not possible for general coefficients.

- #8

tiny-tim

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- #9

fzero

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Hmm, actually this is incorrect. Rotating about the origin does not correspond to subtracting an angle from t.

Of course it does. The rotation matrix is

[tex]\begin{pmatrix} \cos\theta & \sin\theta \\ - \sin\theta & \cos\theta \end{pmatrix}[/tex]

and it acts on

[tex]\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \cos t \\ \sin t \end{pmatrix} .[/tex]

You can verify the action on [tex]t[/tex]. Remember that we have a unit circle based at the origin. What other action would a rotation around the origin have?

To clarify, I tried plugging in t=a+b and finding an equation for b such that only cos(a) and cos^2(a) terms remain. However, this is not possible for general coefficients.

I think the problem you're having is that the center of the 2nd ellipse is not invariant under this transformation.

I just tested my formula for 2 cases. The first was the unit circle and the circle [tex](x-1)^2 + (y-1)^2 =1[/tex], which intersect at two points. The second was the unit circle and the circle [tex](x-1)^2 + (y-1)^2 =(1-\sqrt{2}/2)^2,[/tex] which touch at a point.

You should try to convince yourself that the problem simplifies drastically if either [tex]x_0'[/tex] or [tex]y_0'[/tex] are zero. The rotation I made was a simple way to achieve this.

- #10

Gerenuk

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Actually I don't need the intersections, but I rather only want to know if there are any. So maybe there is a way to solve it.You can't beat the system!

@fzero:

If you just plug in this transformation, you don't get rid of y_0.

As I said, if you do it purely algebraically by setting t=a+b and trying to pick b as to eliminate sin(a) terms, you find it's impossible unless you are dealing with two circles on a diagonal.

So you incidently managed to pick exactly the two special cases where it works :) Tests should be more arbitrary.

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