A Puzzle: Find an Ellipse Tangent to a Circle

[quantumdude]

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!

 

[Bystander]

If m_n = 1 are you in business?

 

[quantumdude]

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]

Yet, m_n = 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]

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.

 

[quantumdude]

$m_n=1$ 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]

Lost the tangent point subscript translating between equation editors. Back to the drawing board.