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Having three given points in the two-dimensional plane and semi-axis lengths $a$ and $b$ of an ellipse, how to determine the center? By construction (the "Euclidean way") or analytic geometry.

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Are you sure you mean axis lengths and not semiaxis lengths? That's what $a$ and $b$ usually denote for an ellipse. – joriki Oct 14 '12 at 19:56
OK, I edited that. $a$ and $b$ are the semi-axes lengths. Anyway, those are given, in the final solution it will be two times less or more. – simco Oct 14 '12 at 20:01
$$b^2\left(c(x_i-M_x) - s(y_i - M_y)\right)^2 + a^2\left(s(x_i-M_x) + c(y - M_y)\right)^2 = a^2b^2$$ for $i\in\{1,2,3\}$ together with $s^2+c^2=1$ gives four equations in four variables ($M_x, M_y, s, c$), but so non-linear that my CAS won't solve them out of the box. So phrasing the problem for analytic geometry sounds fine, but actually solving it that way still requires work. – MvG Oct 15 '12 at 11:15
Thank you for your effort. You have a typo, your last $y$ should be $y_i$. What do you mean by "my CAS"? – simco Oct 15 '12 at 11:35
Yeah, noticed the typo myself, but that was after the grace period, so I couldn't edit my comment any more. CAS = Computer Algebra System. In my case that's maxima, as a backend to sage. – MvG Oct 15 '12 at 11:38
up vote 3 down vote accepted

The equation of an axis-parallel ellipse with semiaxes $a$ and $b$ is

$$ \left(\frac xa\right)^2+\left(\frac yb\right)^2=1\;. $$

Thus, the equation of an arbitrary ellipse with centre at $(x_0,y_0)$ rotated by $\phi$ is

$$ \left(\frac{x\cos\phi-y\sin\phi -x_0}a\right)^2+\left(\frac{x\sin\phi+y\cos\phi-y_0}b\right)^2=1\;. $$

Subsituting your three points into this equation yields three equations for the three unknowns $x_0,y_0,\phi$. Subtracting two pairs of these from each other eliminates the constant terms quadratic in $x_0$ and $y_0$ and results in two equations, still with all three unknowns but now linear in $x_0$ and $y_0$. You can use these to express $x_0$ and $y_0$ solely in terms of $\phi$, namely as fractions of a homogeneous trigonometric polynomial of degree $3$ over one of degree $2$. Substituting these into the above equation and multiplying through by the square of their common denominator (which is the determinant of the $2\times2$ linear system for $x_0$ and $y_0$) yields a homogeneous trigonometic polynomial of degree $6$. This can be rewritten in terms of $\cos2\phi$ and $\sin2\phi$, and then taking the terms with a factor of $\sin2\phi$ to one side, squaring and using $\sin^22\phi=1-\cos^22\phi$ yields an algebraic equation of degree $6$ for $\cos2\phi$ that I think you'll have to solve numerically. You should check which of the resulting values of $\phi$ satisfies the unsquared equation, since the squaring may have introduced spurious solutions. You could also avoid the squaring by solving the trigonometric equation instead of the algebraic one; in that case you'll get two values of $\phi$ that differ by $\pi$ for each possible ellipse. Once you know all valid values of $\phi$, you can substitute them into the $2\times2$ system and solve for the corresponding values of $x_0$ and $y_0$.

You can see from a simple example that it makes sense that there can be up to six solutions: If the three points form an equilateral triangle and the semiaxes are such that the ellipse just fits between two of them with its tip at the third, with a little room to spare, then you can push it past the tip on either side to make it touch the two points; there are three ways to choose the points and two sides of the tip to move to, for a total of six different ways that you can get the ellipse to touch the points.

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Tnx joriki! Does this mean that we might have 3 solutions in some case for the three points? – simco Oct 18 '12 at 21:08
@simco: Sorry, there was an error; I've fixed it; you actually get not a cubic for $\cos^22\phi$ but a full sixth-order equation for $\cos2\phi$, which I think you'll have to solve numerically. I added a paragraph showing that there actually are cases in which all six solutions are valid. – joriki Oct 19 '12 at 10:28

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