# Solve that without 4th order polynomial?

#### Gerenuk

I was trying to calculate when two ellipses of the form ((x-x0)/a)^2+((y-y0)/b)^2=1 have intersections. Most of the time I get 4th order equations. Actually I only want to know a condition IF they just touch. Any ideas on this?

After playing around I had various equations. For example the problem is equivalent to finding beta in
$$\Re\left((\exp(i\beta)+z)^2\right)=b$$
where z is complex. Suggestions?

#### tiny-tim

Homework Helper
Hi Gerenuk!

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?

#### Gerenuk

Yes, I started like this too. Yet I haven't managed to avoid the 4th order equation. I tried setting the discriminant to zero to determine if the ellipses touch, but there were just too many terms, and I could find an easy way to factorize.
The above equation looks the simplest I could find. Yet not sure how to proceed.

#### fzero

Homework Helper
Gold Member
The intersection of two generic conics will always be equivalent to solving the quartic, essentially due to Bezout's theorem. The best that you can do is reduce this to solving a cubic using some algebraic geometry techniques. The method is described in the context of conics at http://en.wikipedia.org/wiki/Conic_section#Intersecting_two_conics and in the context of the quartic at http://en.wikipedia.org/wiki/Quartic_equation#Algebraic_geometry

With two ellipses, it seems like we can simplify things quite a bit. Use linear transformations to map one of the ellipses to a unit circle. This transforms the parameters of the other ellipse, but it's still of the same form that you wrote down before. Let the center of this ellipse be $$(x_0',y_0')$$. Now since the unit circle is invariant under rotations in the plane, we can use a rotation to set, say, $$y_0'=0$$. But then the quartic equation for the intersection points is actually a perfect square so the problem reduces to solving a quadratic equation. The four points of intersection are $$(r_{\pm}, \pm \sqrt{1-r_\pm^2})$$, where $$r_\pm$$ are the roots to the quadratic.

For the ellipses to only touch, these four points must be coincident. This can only be the case if $$r_+=r_- = \pm 1$$. 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.

#### Gerenuk

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

#### fzero

Homework Helper
Gold Member
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 ($$x=\cost t, y=\sin t$$), the equation for the points of intersection is

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

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

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

where

$$x_0''=x_0'\sqrt{ 1+ \left(\frac{y_0'}{x_0'}\right)^2}.$$

The whole point is that (*) is a quadratic equation for $$cos(t-\theta)$$.

#### Gerenuk

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

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.

#### tiny-tim

Homework Helper
Yet I haven't managed to avoid the 4th order equation.
You can't beat the system!

#### fzero

Homework Helper
Gold Member
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

$$\begin{pmatrix} \cos\theta & \sin\theta \\ - \sin\theta & \cos\theta \end{pmatrix}$$

and it acts on

$$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \cos t \\ \sin t \end{pmatrix} .$$

You can verify the action on $$t$$. 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 $$(x-1)^2 + (y-1)^2 =1$$, which intersect at two points. The second was the unit circle and the circle $$(x-1)^2 + (y-1)^2 =(1-\sqrt{2}/2)^2,$$ which touch at a point.

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

#### Gerenuk

You can't beat the system!
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.

@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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