How to draw an ellipse if a center and 3 arbitrary points on it are given?

How to draw an ellipse if a center and 3 arbitrary points on it are given?

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@RahulNarain: Updated. –  kiss my armpit Mar 23 '13 at 22:05

You must find three unknowns - two half axes describing the ellipse and an angle to describe it's orientation.

You must choose some equation that describes ellipse and write the equations for each of the points. Find the parameters of ellipse from them.

For example, with parametric equations you have two equations at each point: $$x_i=x_c+a\cos t_i \cos \varphi - b \sin t_i \sin \varphi$$ $$y_i=y_c+a\cos t_i \sin \varphi - b \sin t_i \cos \varphi$$

where $x_i$ and $y_i$ are the coordinates of $i$-th point, $x_c$ and $y_c$ are the centers coordinates, $t_i$ is parameters $t$ value at the $i$-th point and $a$, $b$ and $\varphi$ are the unknown values describing the ellipse and it's orientation. If you write these equations, you will have six equations and six unknown values (three $t_i$'s and the wanted values). Solvable, but not trivial. I'm showing you the approach not the solution. I'll leave it up to you to choose more appropriate ellipse equations to start with :)

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If a center point $C$ is chosen for the ellipse, one can draw two intersecting ellipses with center at $C$ which intersect at four points. So if $A,B$ happen to be two of those four points, the ellipse will not be unique. So I think you might need more assumptions, maybe that the ellipse is to have its major axis parallel to the $x$ axis.

Just saw the update: The above example shows you need at least 5 points given on the ellipse to determine it.

ADDED: Consider the following two ellipses, both centered at $(0,0)$. $$E_1:\ x^2+2y^2=3,\\ E_2:\ 2x^2+y^2=3.$$ The four points $$(1,1),\ (1,-1),\ (-1,1),\ (-1,-1)$$ each lie on both ellipses $E_1$ and $E_2$. So in general given only four points and the center, the ellipse is not determined.

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5 points? Ellipse with fixed center has exactly three degrees of freedom, so three constraints must be exactly as much as needed. –  Juris Mar 23 '13 at 22:40
@Juris : See the example I inserted after "ADDED" in above answer. –  coffeemath Mar 23 '13 at 23:26

For the ellipse is defined in a unique way, it takes one beyond the three-point line $ax + by + c = 0$ with the eccentricity $e$ the ellipse (see the definition of elipse in see wikipedia for more).

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