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.

  • $\begingroup$ @Peter: But the subsequent equations 21 and 22 make it seem difficult to integrate the given information about the semi-axes into that approach. $\endgroup$ – joriki Jul 22 '11 at 14:21
  • $\begingroup$ I was hoping this would involve using an eccentric anomaly but seems you guys have solved without :) $\endgroup$ – Alice Jul 22 '11 at 16:59

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.


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

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


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
  • $\begingroup$ Wow! Thanks. Will try and understand this. +1 for now. $\endgroup$ – AppleGrew Jul 22 '11 at 20:21
  • $\begingroup$ Here's the x-coordinate ;-) $\frac{1}{2 a b}\left(a b (x_1+x_2)+\left(a^2 (-y_1+y_2) \cos[\alpha]^2+(a-b) (a+b) (x_1-x_2) \cos[\alpha] \sin[\alpha]+b^2 (-y_1+y_2) \sin[\alpha]^2\right) \surd \left(\left(b^2 \left((x_1-x_2)^2+(y_1-y_2)^2\right)+a^2 \left(-8 b^2+(x_1-x_2)^2+(y_1-y_2)^2\right)+(a-b) (a+b) (-(x_1-x_2+y_1-y_2) (x_1-x_2-y_1+y_2) \cos[2 \alpha]-2 (x_1-x_2) (y_1-y_2) \sin[2 \alpha])\right)/\left(-\left(a^2+b^2\right) \left((x_1-x_2)^2+(y_1-y_2)^2\right)+(a-b) (a+b) ((x_1-x_2+y_1-y_2) (x_1-x_2-y_1+y_2) \cos[2 \alpha]+2 (x_1-x_2) (y_1-y_2) \sin[2 \alpha])\right)\right)\right)$ $\endgroup$ – whuber Jul 22 '11 at 21:54
  • $\begingroup$ are you sure this equation is correct? I am not getting the correct results. And I am guessing $\sin[\alpha]^2$ means sine of square of alpha. $\endgroup$ – AppleGrew Jul 25 '11 at 2:40
  • $\begingroup$ You should check jsfiddle.net/6Wxbg . The two pink points are the given points. Ideally the two ellipse must intersect at the two points. $\endgroup$ – AppleGrew Jul 25 '11 at 4:18
  • $\begingroup$ @J.M. "Here's" where? $\endgroup$ – AppleGrew Jul 25 '11 at 4:43

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


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


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.

  • 5
    $\begingroup$ +1 Even easier: after the rotation, rescale the axes by $1/a$ and $1/b$. This reduces the problem to finding the center of a unit circle given two points on it, which is a standard (easy) Euclidean geometry construction. $\endgroup$ – whuber Jul 22 '11 at 15:10
  • $\begingroup$ Ah, yes, that makes sense :-) $\endgroup$ – joriki Jul 22 '11 at 15:12
  • 3
    $\begingroup$ Indeed, you can then further rotate and translate the configuration to place the two points at $(0,\pm c)$. Clearly then the solutions for the center are $(\pm s, 0)$ where $s^2=1-c^2$. (Obviously there's no solution when $|c| \gt 1$.) $\endgroup$ – whuber Jul 22 '11 at 15:21
  • $\begingroup$ You should write that as an answer; it's better than mine :-) $\endgroup$ – joriki Jul 22 '11 at 15:26
  • $\begingroup$ It's the same as yours; these are just comments :-). $\endgroup$ – whuber Jul 22 '11 at 15:58

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

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

  • $\begingroup$ Well checkout jsfiddle.net/6Wxbg/1 . As I mentioned this is not giving me the desired result. The two ellipses should intersect at the pink dots. $\endgroup$ – AppleGrew Jul 25 '11 at 5:04
  • $\begingroup$ I don't understand why there is "the x-coordinate" -- shouldn't there be two x-coordinates, one for each centre? The green ellipse seems to be on target (it hits the top left corners of the pink dots, which is correct since you draw those from there), so this x-coordinate seems to be the correct one for cy2, and you probably need to change a sign somewhere to make it work for cy1 -- you can sort of tell that the red ellipse will be OK if you slide it to the right. $\endgroup$ – joriki Jul 25 '11 at 5:30
  • $\begingroup$ I haven't checked whuber's solution in detail (yet) so I can't say anything... $\endgroup$ – J. M. isn't a mathematician Jul 25 '11 at 5:32
  • 3
    $\begingroup$ @All I could provide the y-coordinate of this solution, along with (x,y) for the second solution, but that's not the point. Isn't it obvious that implementing such lengthy convoluted formulas is inferior to implementing the algorithm as described? But if anybody wants the full solution, just ask Mathematica to carry out a FullSimplify of the generic formula FullSimplify[f[\[Alpha], a, b, {x1, y1}, {x2, y2}] (applying suitable domain restrictions and after removing the If statements to check for special cases.) $\endgroup$ – whuber Jul 25 '11 at 13:51
  • 1
    $\begingroup$ @J.M. Thanks for introducing me to the align* construct--it's just what is needed. Incidentally, I tested the Mathematica code (for the function f which produced this formula) by creating a lot of random configurations, applying the code, and checking that it returned the correct center every time. Although something may have happened in the process of copying the formula from Mathematica and reformatting it here--which further illustrates the dangers of implementing long complex formulas in code--, I can vouch for the correctness of the overall approach. $\endgroup$ – whuber Jul 25 '11 at 21:59

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.

  • $\begingroup$ Thank you for pointing that out. My exposition indeed was off by a factor of $1/2$--but the code was correct, and no other part of the solution was affected. $\endgroup$ – whuber Jun 10 '16 at 13:05

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