Common tangent ellipse and circle

kingofinismac · #1 $Old$ June 10th, 2009, 06:03 AM

hi guys,

I'm working on a fairly complex problem and it has come down to one final part.

I have an overlapping circle and ellipse. I have the equation for both.

The centre of the circle is say (5,5) and the foci of the ellipse are say (5,5) and (10,5), so one of the foci is the centre of the circle.

I need to find the equation of the 2 common tangents (and consequently, the intersection points of the tangents with the circle and ellipse).

It seems fairly straightforward but I'm having difficulty getting it out.

Any help greatly appreciated!

M

alunw · #2 $Old$ June 10th, 2009, 08:36 AM

This may not be much help to you, but have you considered changing coordinates so that the centre of the circle is at 0,0. Then at a point (a,b) on the circle you have aa+bb=rr
where r is the radius of the circle and the equation of the tangent there is ax+by-rr=0.
You will probably also want to put a=r cos theta and b = r sin theta as you imagine yourself traveling round the circle and theta goes from 0 to two*pi radians, and then maybe use the half angle formulas for sin and cos in terms of t = tan(theta/2) to get rid of the trig functions.
Then you can substitute those values into the (new) equation of the ellipse to find the intersections of the circle tangent at (a,b) with the ellipse. That should get you a quadratic equation in t, and the common tangents will be the points where the quadratic has equal real roots, so you can use the usual formula for solving quadratics and find where the bit you take the square root of is zero.
You could probably do things the other way around, and that might be easier but this way you don't have to know the general rule for the equation of a tangent to an ellipse.

kingofinismac · #3 $Old$ June 10th, 2009, 12:44 PM

Thanks alunw,

That's an interesting approach and not one I had considered.

I understand it apart from the use of half angle formulas to get rid of the trig functions. Could you clarify this please?

Thanks,
M

alunw · #4 $Old$ June 10th, 2009, 05:49 PM

Equations with cos and sin in them are going to be awkward.
So you can get rid of them using the formulas

sin(theta) = 2*t/(1+t^2) cos(theta)=(1-t^2)/(1+t^2) where t=sin(0.5*theta)/cos(0.5*theta) (i.e. tan(0.5*theta))

I'm slightly worried that my suggestion is probably not the best possible. There is probably some neat geometric trick using some property of a tangent or an ellipse, but the only property I know for an ellipse is that the sum of the distances to a point on an ellipse from the two foci is constant, and I couldn't see how this would help.

pankaj · #5 $Old$ June 10th, 2009, 07:49 PM

There is another very simple approach when two degree curves are involved.

Let $y=mx+c$ be the equation of line which touches both the curves.
Solve for point of intersection with the given two curves.

This results in obtaining two quadratics.Since the line is tangent to both the curves, both the quadratics must give one solution each because of which the discriminants of the two quadratics must be zero.

This will give you two equations in $c$ and $m$ .
Now solve for $c$ and $m$ .

You forgot to mention thr radius of the circle in your question.

Try these questions.Find equation of line which touches both the curves
$1)y^2=8x ;xy =-1$ .
$2)y^2=4ax ;x^2=32ay$
$3)y=x^2 ; y=-x^2+4x-4$
$4)x^2+y^2=2a^2 ; y^2=8ax$

This method works.

kingofinismac · #6 $Old$ June 11th, 2009, 04:06 AM

pankaj, thanks for the help!

If I consider your first example.

I first find the intersection of the line with the first curve; that gives me an expression for x and y in terms of m and c

I then find the intersection of the line with the second curve; once again that gives me an expression for x and y in terms of m and c.

So now I have four equations corresponding to x1,y1 and x2, y2, the two intersection points. From here, I don't really understand how you solve for m and c?

Thanks,
M

alunw · #7 $Old$ June 11th, 2009, 04:47 AM

I don't think this is really very different from what I suggested. All I have done is restrict consideration to lines that are tangent to the circle, and then suggested a change of variable that I think will make the problem easier to solve.

I don't like using y=mx+c as the equation of a straight line since it is no good for vertical lines. A much better equation for lines is ax+by+c=0 where you arrange for aa+bb to be 1. (The equation I gave for the tangent of the circle is in this form if you divide by the radius). This works for any line in the plane, and for points off the line the quantity ax+by+c is then the signed distance from the line (so you can tell what side of the line you are on) and c is the distance from the line of the origin. And problems involving intersections are often easiest if you use parametric equations x=f(t),y=g(t) and solve for t. There is usually less trouble with square roots than if you try to solve the quadratic by eliminating one of x and y.

pankaj · #8 $Old$ June 11th, 2009, 06:05 AM

O.K..I will solve the first one for you and I am sure you will be able to pick up from there.

Let $y=mx +c$ be the required line.

Solving with $y^2=8x$ yields

$(mx+c)^2=8x$

$m^2x^2+2(mc-4)x+c^2=0$

which is a quadratic in $x$ which must have $one$ root.

$Discriminant=0$

$(mc-4)^2-m^2c^2=0$

$c=\frac{2}{m}$ .....................(1)

Now solving $y=mx+c$ and $xy=-1$ yields

$x(mx+c)=-1$

$mx^2+cx+1=0$

which is a quadratic in $x$ which must have one root.

$Discriminant=0$

$c^2-4m=0$

$c^2=4m$ ..........................(2)

Solving (1) and (2) we get

$m=1$ and $c=2$

Therefore,the equation of the common tangent is $y=x+2$ .

Apply the same method to other problems

kingofinismac · #9 $Old$ June 12th, 2009, 01:20 AM

Thanks for the help pankaj!

I've tried that method on a specific example I'm working on. The two equations are the circle:
$X^2 + Y^2 = 1$ and

and ellipse:
$((X-2)^2)/3 + Y^2 = 1$

The first equation in terms of m and c (by setting the determinant equal to zero) is
$m^2 -c^2 +1 = 0$

The second equation in terms of m and c (by setting the determinant equal to zero) is
$12-48mc - 12c^2 - 12m^2 = 0$

Solving these equations yields the following expression for c:
$3c^4 - 2c^2 -1 = 0$

Now when I solve this in maple, and plot the resultant line, it does not seem to be tangential to the two curves.

Am I doing something wrong?

Thanks,
M

pankaj · #10 $Old$ June 12th, 2009, 02:09 AM

$(3c^2+1)(c^2-1)=0$

$c=\pm1$

$m=0$

This means equations of tangents are $y=1$ and $y=-1$

These 2 lines are parallel to x-axis and are touching both the curves

kingofinismac · #11 $Old$ June 12th, 2009, 03:23 AM

Perfect, thanks mate!

kingofinismac · #12 $Old$ June 27th, 2009, 12:43 PM

hey guys, I'm afraid I'm back again.

I've done the maths and written some code.

I've a particular example that I'm having a bit of a problem with:

I've a circle and an ellipse. I solve for c using the method suggested above and that yields the following:

-1.4220
-1.0230
1.4220
1.0230

This makes sense since we'd expect 4 tangents.

But I solve for m as follows:
m = plus or minus sqrt(c^2 - 1)

so for c = -0.142, I'll have two values of m, and therefore 2 lines.
and by following the procedure for each value of c, I'll end up with 8 lines.

Why do I have 4 extra lines and how can I get rid of the ones I don't need?

Thanks,
M

pankaj · #13 $Old$ June 27th, 2009, 06:00 PM

Tell me the exact question.What are the equations of circle and ellipse

kingofinismac · #14 $Old$ June 28th, 2009, 02:02 AM

Ellipse: ((x-5)^2/(9)) + (((y)^2)/(4))=1

Circle: x^2 + y^2 =1

The issue is I do get 4 correct tangents, but also these extra lines.

Thanks,
M

pankaj · #15 $Old$ June 28th, 2009, 01:31 PM

If $x^2+y^2=a^2$ and if $y=mx+c$ is a tangent then $c^2=a^2(1+m^2)$

Therefore any tangent to the circle can be written as $y=mx\pm a\sqrt{1+m^2}$

Similarly,if $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is an ellipse and $y=mx+c$ be tangent to it then $c^2=a^2m^2+b^2$

Therefore any tangent to this ellipse can be expressed as $y=mx\pm \sqrt{a^2m^2+b^2}$

But if ellipse is $\frac{(x-5)^2}{a^2}+\frac{y^2}{b^2}=1$

then making corresponding change in equation of tangent,we get the equation of tangent to be

$y=m(x-5)\pm \sqrt{a^2m^2+b^2}$

Therefore the equation of tangent to $x^2 +y^2=1$ is $y=mx\pm \sqrt{1+m^2}$

And equation of tangent to $\frac{(x-5)^2}{9}+\frac{y^2}{4}=1$ is $y=m(x-5)\pm \sqrt{9m^2+4}$ .

If the two equations of tangents represent the same line then
$\pm \sqrt{1+m^2}=-5m\pm \sqrt{9m^2+4}$

$\mp \sqrt{9m^2+4} \pm \sqrt{1+m^2}=-5m$

Now herein lies answer to your puzzle.

To solve the equation you obviously square both the sides.On squaring you not only get the solution of above equation but you have also unknowingly squared the equation
$\mp \sqrt{9m^2+4} \pm \sqrt{1+m^2}=5m$

(note the RHS of both the sides)

So the four extra solutions that you are getting are the solutions of the equation $\mp \sqrt{9m^2+4} \pm \sqrt{1+m^2}=5m$

and NOT of $\mp \sqrt{9m^2+4} \pm \sqrt{1+m^2}=-5m$ which you intended to solve in the first place.

I am afraid there is no other way but to check manually whether the solutions you obtain satisfy the original equation or not.