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Tom Mattson
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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{1e^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:
$$\frac{x^2}{a^2}+\frac{y^2}{d^2a^2}=1$$
$$(xh)^2+(yk)^2=R_i^2$$
$$y=m_n(xh)+k$$
$$y=xa+da$$
Here are the relationships that I have derived so far.
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{1e^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:
$$\frac{x^2}{a^2}+\frac{y^2}{d^2a^2}=1$$
$$(xh)^2+(yk)^2=R_i^2$$
$$y=m_n(xh)+k$$
$$y=xa+da$$
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=aR_o## and ##k=daR_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_0k}{x_0h}## 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.
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