# A Puzzle: Find an Ellipse Tangent to a Circle

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I was asked to look at a problem (not homework) in which a tangent ellipse is to be found for a circle. This puzzle is turning out to be more than I bargained for. See the attached image because hey, a picture's worth a thousand words.

The givens in this problem are to be the radius ##R_i## of the circle, the eccentricity ##e## of the ellipse, and the distance ##R_o## from the center of the circle to either of the lines tangent to the ellipse. I also have that the tangent ellipse is to be centered at the origin and that it has a horizontal major axis.

I am to find the semiaxes ##a## and ##b## of the ellipse, the center ##(h,k)## of the circle, the point ##\left(x_0,y_0\right)## of tangency, and the slope ##m_t## of the circle/ellipse at the point of tangency. The horizontal and vertical tangents to the ellipse are ##y=b## and ##x=a##, respectively, but I don't know exactly where they are.

I have developed a little bit of shorthand to make the equations simpler. I've let ##d=\sqrt{1-e^2}## and ##m_n=-1/m_t##. The latter is the slope of the normal line that passes through ##\left(x_0,y_0\right)##. I also noted the line that passes through ##(h,k)## and ##(a,b)## has slope 1, so I thought it could be of interest. I call it "the 45 degree line". So the equations of the ellipse, circle, normal line, and 45 degree line are, respectively:

$$\frac{x^2}{a^2}+\frac{y^2}{d^2a^2}=1$$
$$(x-h)^2+(y-k)^2=R_i^2$$
$$y=m_n(x-h)+k$$
$$y=x-a+da$$

Here are the relationships that I have derived so far.

• Because the center of the circle is to be the same distance from either of the tangent lines ##y=b## or ##x=a##, I have ##h=a-R_o## and ##k=da-R_o##.
• Using the equation of the ellipse and its implicit derivative, I have ##x_0=\frac{a}{\sqrt{1+m_n^2d^2}}## and ##y_0=\frac{m_nd^2a}{\sqrt{1+m_n^2d^2}}##.
• Plugging the last two relations into ##m_n=\frac{y_0-k}{x_0-h}## I obtained a 4th degree polynomial equation for ##m_n##. Only one problem: it also contains the unknown quantity ##a##. I'll post the equation if anyone really wants to see it, but it's really not necessary. Solving that equation would be the final step, and I would throw it into Maple anyway.
I feel like I'm on the verge of solving this problem. If I could just find any of the unknowns ##a,b,h,## or ##k##, I could find all of the others and this thing would be done. But I've been chasing my own tail for so long that I just can't see my way out of it. I'm hoping that a fresh pair of eyes will help. Thanks!

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Bystander
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If mn = 1 are you in business?

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Yes, and I considered that too. If ##m_n=1## then the normal line *is* the 45-degree line, which certainly makes the math easier. However, if that's the case then there isn't a solution for every choice of ##e,R_i##, and ##R_o##. In fact I don't know if there's a solution for any choice of those variables if ##m_n=1##.

Bystander
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Yet, mn = 1 drops out of what you've posted. Can't say I'm comfy with the implications, and wasn't up all night mulling the geometry, but it's got me going just the same.

Bystander
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there isn't a solution for every choice of e,R i e,R_i, and R o R_o.
I've been confining the problem statement to a "given" circle with a "given" center, which it is not. You get to move the circle, so there is a range of eccentricity that is possible.

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##m_n=1## does not drop out of those equations. What does drop out is this:

##d^2\left(a-R_o\right)^2m_n^4-2d^2\left(a-R_o\right)\left(da-R_o\right)m_n^3+\left(\left(a-R_o\right)^2+d^2\left(da-R_o\right)^2-e^4a^2\right)m_n^2-2\left(a-R_o\right)\left(da-R_o\right)m_n##
##+\left(da-R_o\right)^2=0##

If I had ##a## (or ##h,k,## or ##b##, which would be just as good), I could evaluate the coefficients of this equation and chuck it into Maple to solve for ##m_n##. Eccentricity is one of the givens. That's what is making this so hard. If ##a## and ##b## were both given seperately, this would be an exercise in freshman calculus.

Bystander