Consider an ellipse with semi-axes $a$ and $b$, taller than it is wide with a small circle of radius $r$ inside. Assume the circle falls to the lowest point possible while staying inside the ellipse.

If $2r\le a-c$ then the circle and ellipse will meet at a single point at the bottom. If $2r>a-c$ the circle and ellipse will intersect at two points on the opposite side, leaving a space between the bottom of the circle and the bottom of the ellipse. For this case, given the radius of the circle and the dimensions of the ellipse how do I calculate the distance $d$ between the bottom of the circle and the bottom of the ellipse?


  • $\begingroup$ What is c? The semi-axes are a and b and the circle radius is r. Is this correct? $\endgroup$ – ja72 Oct 17 '16 at 18:43
  • $\begingroup$ @ja72 2c is the distance between two focii $\endgroup$ – iamgr007 Oct 17 '16 at 19:04
  • $\begingroup$ The circle will only have one contact point if the radius is less than the minimum radius of curvature which equals to $\frac{a^2}{b}$. $\endgroup$ – ja72 Oct 17 '16 at 19:50
  • $\begingroup$ With an horizontal scalling the ellipse may be transformed in a circle and we get this (slightly more general) treatment in accord with the fine answers here. $\endgroup$ – Raymond Manzoni Oct 18 '16 at 12:01

I got $$w =b - \sqrt{\frac{(b^2-a^2)(a^2-r^2)}{a^2} }-r $$

Start with polar coordinates for the ellipse $$ \begin{pmatrix}x \\ y \end{pmatrix} = \tfrac{a b}{\sqrt{b^2-(b^2-a^2) \cos^2 \varphi}} \begin{pmatrix} \sin \varphi \\ -\cos\varphi \end{pmatrix}$$ where $(x,y)$ are the contact point coordinates relative to the center of the ellipse

The contact angle $\eta$ (see below) is found from the tangent vector of the ellipse as $$ \eta = \varphi - \tan^{-1} \left( \frac{ (a^2-b^2) \sin\varphi \cos\varphi}{b^2 - (b^2-a^2)\cos^2 \varphi} \right) $$ pic

the location of the circle is found from the relationships $$\begin{aligned} r \sin \eta = s \sin \varphi \\ z+r \cos \eta = s \cos \varphi \end{aligned}$$

This is solved for $$\varphi = \frac{\pi}{2} - \tan^{-1} \left( \frac{b^2}{a} \sqrt{ \frac{a^2-r^2}{b^2 r^2-a^4} } \right) $$

and $$ z = s \cos\varphi - r \cos \eta $$

Finally, the gap is $$\boxed{w = b -z - r}$$ and with a lot of simplifications (thank you CAS) I got the answer above.

  • $\begingroup$ Not that you can also arrive at this using the radius of curvature (equal to circle) $$ \frac{1}{r}=a b \frac{\left(b^{2}-(b^{2}-a^{2})\cos^{2}\varphi\right)^{\frac{3}{2}}}{\left(b^{4}-(b^{4}-a^{4})\cos^{2}\varphi\right)^{\frac{3}{2}}} $$ $\endgroup$ – ja72 Oct 17 '16 at 19:55
  • $\begingroup$ If you swap $a$ and $b$ so that $a$ is the semi-major axis and $b$ is the semi-minor axis, then your answer matches mine. I was getting bizarre results plugging in the $a$, $b$, and $r$ I had for my test ellipse. $\endgroup$ – robjohn Oct 18 '16 at 11:34
  • $\begingroup$ Sorry, I must have used a different convention. $\endgroup$ – ja72 Oct 18 '16 at 13:21
  • $\begingroup$ cartesian coordinates gives a much quicker result, with much less basic knowledge, and no CAS help necessary. $\endgroup$ – dfnu Jan 24 at 15:08

enter image description here

By the Law of Sines, we have that $$ \frac{|CF_1|}{|PF_1|}=\frac{\sin(\alpha)}{\sin(\theta)}=\frac{\sin(\alpha)}{\sin(\pi-\theta)}=\frac{|CF_2|}{|PF_2|}\tag{1} $$ and by the defining property of an ellipse, $$ \frac{|CF_1|+|CF_2|}{|PF_1|+|PF_2|}=e=\frac{\sqrt{a^2-b^2}}{a}\tag{2} $$ Thus, $(1)$ and $(2)$ imply $$ \frac{|CF_1|}{|PF_1|}=\frac{|CF_2|}{|PF_2|}=e\tag{3} $$ The Law of Cosines says that $$ |PC|^2+|PF_1|^2-2|PC||PF_1|\cos(\alpha)=e^2|PF_1|^2\tag{4} $$ and $$ |PC|^2+|PF_2|^2-2|PC||PF_2|\cos(\alpha)=e^2|PF_2|^2\tag{5} $$ Therefore, since $|PF_1|+|PF_2|=2a$, subtracting $(5)$ from $(4)$ and dividing by $|PF_1|-|PF_2|$ yields $$ \frac{b^2}a=a\left(1-e^2\right)=|PC|\cos(\alpha)\tag{6} $$ Plugging $(6)$ into $(4)$ and solving for $|PF_1|$ gives $$ |PF_1|=a\left(1-\sqrt{1-\frac{|PC|^2}{b^2}}\right)\tag{7} $$ and therefore, $$ |CF_1|=\sqrt{a^2-b^2}\left(1-\sqrt{1-\frac{|PC|^2}{b^2}}\right)\tag{8} $$ Letting $r=|PC|$, the distance from the right end of the circle to the right end of the ellipse is $|CF_1|+a-\sqrt{a^2-b^2}-r$, that is $$ \bbox[5px,border:2px solid #C0A000]{d=a-r-\sqrt{a^2-b^2}\sqrt{1-\frac{r^2}{b^2}}}\tag{9} $$ for $\frac{b^2}a\le r\le b$ and $d=0$ for $r\le\frac{b^2}a$.


Never underestimate the power of cartesian coordinates

Consider the problem of finding the intersection points between ellipse and circle of radius $r$ centered in $(0,y_c)$, which leads to the system of equations $$\begin{cases}\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1\\ x^2 + (y-y_c)^2 = r^2.\end{cases} $$ By substitution you obtain the following quadratic equation. $$ y^2\left(\frac{1}{b^2} -\frac{1}{a^2} \right) + 2y\left(\frac{y_c}{a^2}\right) + \frac{r^2-y_c^2-a^2}{a^2}=0. $$ Since you want the two curves to intersect at only two points with the same ordinate, than this equation must have discriminant equal to $0$, which gives you \begin{eqnarray} \left(\frac{y_c}{a^2}\right)^2-\left(\frac{1}{b^2} -\frac{1}{a^2} \right)\left(\frac{r^2-y_c^2-a^2}{a^2}\right) = 0. \tag{1}\label{discr} \end{eqnarray} Now you just need to solve the quadratic equation \eqref{discr} with respect to the unknown $y_c$. This rapidly gives $$y_c^2 = \frac{(b^2-a^2)(a^2-r^2)}{a^2}.$$ The desired distance is $$w = b-|y_c| - r,$$ that is

$$ w = b - \frac{\sqrt{(b^2 - a^2)(a^2-r^2)}}{a} -r,$$

as, of course, in the other proposed solutions.

  • $\begingroup$ Are you sure? I am getting $y_c^2 = \frac{(a^2-b^2)(r^2-a^2)}{a^2}$ when I try to solve the two quadratics while matching tangents (slopes). Dang you fixed it before I submitted the comment. $\endgroup$ – ja72 Jan 24 at 15:28
  • $\begingroup$ @ja72, yep, corrected error in text. thanks! $\endgroup$ – dfnu Jan 24 at 15:30
  • $\begingroup$ @ja72 still a sign error, hold on :))) $\endgroup$ – dfnu Jan 24 at 15:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.