# How to find the smallest enclosing ellipse around two circles?

Given two circles (defined by center and radius), how do I find the smallest ellipse that encloses both of them? I.e. I search the green ellipse in the picture below.

The ellipses can be considered axes aligned if this simplifies things. The end goal is to classify points on whether they are in the ellipse or not. So a final representation suitable to the form used in this question will be preferred.

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What do you mean by the smallest? Are you considering the area, or maybe the larger radius or maybe perimeter or yet something else? – dtldarek Mar 6 '13 at 8:42
Are the radii of the two circles the same? – Christian Blatter Mar 6 '13 at 9:06
@ChristianBlatter: They are not necessarily the same. – fho Mar 6 '13 at 9:42
@dtldarek: The one with the smallest minor axis with only two intersection points on the circles. – fho Mar 6 '13 at 9:43
The minor axis doesn't seem to be an interesting quantity to minimize: The minimal possible value of the minor axis is the radius of the larger circle; and when both circles have the same radius, the ellipse will be infinite then.– When you want to minimize the area there will be many cases, depending on the radii of the two circles and the distance of their centers. – Christian Blatter Mar 6 '13 at 20:25

Join the two centers and extend the line. wherever it cuts the two circles at their outer extremes are the points which are crucial say $A$ and $B$. take the mid point of these two points. for simplicity consider this point as the origin.

now consider an ellipse with the major axis as the line segment $AB$ and call this as $x$-axis. you need to find the minor axis and you are done. the equation of an ellipse is

$$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$$

here $a$ is known and $b$ is to be found. find the radius of the osculating circle of this ellipse at the points $A$ and $B$, which is of course the same numerical value (why?)

then equate this numerical value to the maximum of the two radii. this will give you an equation in $b$ which you can solve.

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@Florian The curvature of an ellipse can be found here (search for "The curvature and tangential angle of the ellipse are given by"). – dtldarek Mar 6 '13 at 8:58
I am not sure I understand the "then equate this numerical value" part. Given the equation I can solve for b, use the numerical value as a and one of the intersection points as (x,y). Will this give me the smallest b? Or am I mistaken here? – fho Mar 6 '13 at 10:35
i wanted to give this as an exercise as the computations are a bit cumbersome to write. do u know what an osculating circle is ? to find it you need to consider the distance parametrized curve ... tell me if you do not understand all this i will explain in a new answer – magguu Mar 6 '13 at 15:03
@magguu: I don't :) ... but it would be great if you could point me to some material on it. I have another project that might benefit from this. – fho Mar 7 '13 at 11:12
@Florian There are many excellent books. You may refer to the first chapter of doCarmo, Differential Geometry of Curves and Surfaces. – magguu Mar 7 '13 at 12:11