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.
Best Answer
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 $O_i$:
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).