How to find the center of an ellipse?

I have the following data:-

  • I have two points ($P_1$, $P_2$) that lie somewhere on the ellipse’s circumference.
  • I know the angle ($\alpha$) that the major-axis subtends on x-axis.
  • I have both the radii ($a$ and $b$) of the ellipse.

I now need to find the center of this ellipse. It is known that we can get two possible ellipses using the above data.

I have tried solving this myself but the equation becomes so complex that I always give up.

This is what I have done till now:-

I took the normal ellipse equation $x^2/a^2 + y^2/b^2 = 1$. To compensate for the rotation and translation, I replaced $x$ and $y$ by $x\cos\alpha+y\sin\alpha-h$ and $-x\sin\alpha+y\cos\alpha-k$, respectively. $h$ and $k$ are x and y location of the ellipse’s center.

Using these information I ended up with the following eq:-
$$a B_1\pm\sqrt{a^2 B_1^2 – C_1(b^2 h^2 – 2 A_1 b^2 h)} = a B_2\pm\sqrt{a^2 B_2^2 – C_2 (b^2 h^2 – 2 A_2 b^2 h)} \quad (1)$$

where $A = x\cos\alpha +y\sin\alpha$, $B = -x\sin\alpha+y\cos\alpha$ and $C = a^2 B^2 + A^2 b^2 – a^2 b^2$.

Now the only thing I need to get is $h$ from (1). All other values are known, but I am not able to single that out.

Anyway if the above equations looks insane then please solve it yourself, your way. I could have drifted into some very complicated path.

Solutions Collecting From Web of "How to find the center of an ellipse?"

Let the points be $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$, assumed to lie on an ellipse of semiaxes $a$ and $b$ with the $a$ axis making angle $\alpha$ to the $x$ axis.

Ellipse

Following @joriki, we rotate the points $P_i$ by $-\alpha$ into points

$$Q_i(x_i \cos(\alpha) + y_i \sin(\alpha), y_i \cos(\alpha) – x_i \sin(\alpha)).$$

We then rescale them by $(1/a, 1/b)$ to the points

$$R_i(\frac{x_i \cos(\alpha) + y_i \sin(\alpha)}{a}, \frac{y_i \cos(\alpha) – x_i \sin(\alpha)}{b}).$$

Circle

These operations convert the ellipse into a unit circle and the points form a chord of that circle. Let us now translate the midpoint of the chord to the origin: this is done by subtracting $(R_1 + R_2)/2$ (shown as $M$ in the figure) from each of $R_i$, giving points

$$S_1 = (R_1 – R_2)/2, \quad S_2 = (R_2 – R_1)/2 = -S_1$$

each of length $c$. Half the length of that chord is

$$c = ||(R_1 – R_2)||/2 = ||S_1|| = ||S_2||,$$

which by assumption lies between $0$ and $1$ inclusive. Set

$$s = \sqrt{1-c^2}.$$

The origin of the circle is found by rotating either of the $S_i$ by 90 degrees (in either direction) and rescaling by $s/c$, giving up to two valid solutions $O_1$ and $O_2$. (Rotation of a point $(u,v)$ by 90 degrees sends it either to $(-v,u)$ or $(v,-u)$.) For example, in the preceding figure it is evident that rotation $R_1$ by -90 degrees around $M$ and scaling it by $s/c$ will make it coincide with the circle’s center. Reflecting the center about $M$ (which gives $2M$) produces the other possible solution.

Unwinding all this requires us to do the following to the $O_i$:

  • Translate by $(R_1+R_2)/2$,
  • Scale by $(a,b)$, and
  • Rotate by $\alpha$.

The cases $c \gt 1$, $c = 1$, and $c=0$ have to be treated specially. The first gives no solution, the second a unique solution, and the third infinitely many.

FWIW, here’s a Mathematica 7 function. The arguments p1 and p2 are length-2 lists of numbers (i.e., point coordinates) and the other arguments are numbers. It returns a list of the possible centers (or Null if there are infinitely many).

f[\[Alpha]_, a_, b_, p1_, p2_] := Module[
   {
    r, s, q1, q2, m, t, \[Gamma], u, r1, r2, x, v
    },
   (* Rotate to align the major axis with the x-axis. *)
   r = RotationTransform[-\[Alpha]];
   (* Rescale the ellipse to a unit circle. *)
   s = ScalingTransform[{1/a, 1/b}];
   {q1, q2} = s[r[#]] & /@ {p1, p2};
   (* Compute the half-length of the chord. *)
   \[Gamma] = Norm[q2 - q1]/2;
   (* Take care of special cases. *)
   If[\[Gamma] > 1, Return[{}]];
   If[\[Gamma] == 0, Return[Null]];
   If[\[Gamma] == 1, 
    Return[{InverseFunction[Composition[s, r]][(q1 + q2)/2]}]];
   (* Place the origin between the two points. *)
   t = TranslationTransform[-(q1 + q2)/2];
   (* This ends the transformations.  
   The next steps find the centers. *)
   (* Rotate the points 90 degrees. *)
   u = RotationTransform [\[Pi]/2];
   (* Rescale to obtain the possible centers. *)
   v = ScalingTransform[{1, 1} Sqrt[1 - \[Gamma]^2]/\[Gamma]];
   x = v[u[t[#]]] & /@ {q1, q2};
   (* Back-transform the solutions. *)
   InverseFunction[Composition[t, s, r]] /@ x
   ];

You can make your life a lot easier by rotating the two given points through $-\alpha$ instead of rotating the coordinate system through $\alpha$. Then you can work with the much simpler equation

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

If you substitute your two (rotated) points $(x_1,y_1)$ and $(x_2,y_2)$, you get two equations for the two unknowns $x_0$,$y_0$. Subtracting these from each other eliminates the terms quadratic in the unknowns and yields the linear relationship

$$\frac{x_1^2-x_2^2-2(x_1-x_2)x_0}{a^2}+\frac{y_1^2-y_2^2-2(y_1-y_2)y_0}{b^2}=0\;.$$

You can solve this for one of the unknowns and substitute the result into one of the two quadratic equations, which then becomes a quadratic equation in the other unknown that you can solve.

Of course in the end you have to rotate the centre you find back to the original coordinate system.

Since the expression in whuber’s comment was too darn long, here’s the x-coordinate expression:

$$\begin{align*}
&\frac1{2ab}\left(ab (x_1+x_2)+\left(a^2 (y_2-y_1)\cos^2\alpha+(a-b)(a+b) (x_1-x_2) \cos\,\alpha\sin\,\alpha+b^2 (y_2-y_1)\sin^2\alpha\right)\right.\\
&\left.\surd \left(\left(b^2 \left((x_1-x_2)^2+(y_1-y_2)^2\right)+a^2 \left(-8 b^2+(x_1-x_2)^2+(y_1-y_2)^2\right)+\right.\right.\right.\\
&\left.\left.\left.(a-b)(a+b) (-(x_1-x_2+y_1-y_2) (x_1-x_2-y_1+y_2) \cos\,2\alpha-2 (x_1-x_2) (y_1-y_2) \sin\,2\alpha)\right)/\right.\right.\\
&\left.\left.\left(-\left(a^2+b^2\right) \left((x_1-x_2)^2+(y_1-y_2)^2\right)+(a-b) (a+b) ((x_1-x_2+y_1-y_2) (x_1-x_2-y_1+y_2) \cos\,2\alpha+\right.\right.\right.\\
&\left.\left.\left.2(x_1-x_2)(y_1-y_2)\sin\,2\alpha)\right)\right)\right)
\end{align*}$$

Unfortunately, my correction on the answer of whuber has been rejected, but I’ll try to explain it here.

The chosen answer is very good, but there is an error. $c$ is supposed to be the distance from point $M$ to point $R_1$. The points $R_1$ and $R_2$ are then converted into $S_1$ and $S_2$, respectively, whereby $S_2 = -S_1$. Each $S_1$ and $S_2$ are of length c, as is correctly stated.
So the value of c is actually
$$c = ||S_1 – S_2||/2 = ||2S_1||/2 = ||S_1||$$
apposed to $c \neq ||S_1||/2$.

It occurred to me while trying to implement the proposed solution. Using this correction, the formula works as expected.