# Finding center and rotation angle of ellipse that contains three points

Given three points $$p_1, p_2, p_3 \in \mathbb{R}^2$$, and an ellipse with shape parameters $$(a,b)$$ (the semi-major and semi-minor), is it possible to determine, if they exist, a center $$c \in \mathbb{R}^2$$ and a rotation angle $$\theta \in [0, \pi]$$, such that the ellipse centered at $$c$$ rotated by $$\theta$$ contains $$p_1, p_2, p_3$$?

In other words, let $$E(p, \theta)=\dfrac{(p_x\cos{\theta} + p_{y}\sin{\theta})^2}{a^2} + \dfrac{(p_x\sin{\theta} -p_y\cos{\theta})^2}{b^2}$$

I want to determine $$c\in \mathbb{R}^2$$ and $$\theta \in [0, \pi]$$, such that:

$$$$E(p_1-c, \theta) = 1\\ E(p_2-c, \theta) = 1\\ E(p_3-c, \theta) = 1$$$$

• Is it clear to you that the solution for given $a$ and $b$ may be not possible? Moreover this impossibility will be rather a rule than exception. Knowing escentrity however should be enough for the construction.
– user
May 15, 2019 at 21:49
• One of the terms of $E(p,\theta)$ should have a negative sign to it. May 15, 2019 at 22:28
• Yes, it is clear that there might not be any solution. What do you mean that it should be a rule? Like, I should include it in the problem statement? May 15, 2019 at 23:14
• Welcome to Math.SE. I haven't solve this problem, but I'll provide some hints that may eventually be useful. In the cases where a solution exists, then you can solve it, hopefully, in a step-by-step approach: 1. Use equation for $p_1$ to isolate $c_x$. 2. Substitute $c_x$ in the equation for $p_2$ and solve it for $c_y$. 3. Substitute $c_x$ and $c_y$ in the equation for $p_3$. You will get a large equation with $\sin \theta$ and $\cos \theta$. Hopefully, you may solve that using trigonometric identities. Have you tried this approach? May 16, 2019 at 1:26
• Thanks for the welcome! I tried going in that direction, but isolating $c_x$ involves solving a quadratic equation, which makes things start getting kinda messy. I tried working out the math, but couldn't isolate $c_x$ and $c_y$ as functions of $\theta$. If you can post at least whatever you have done, it would help a lot. May 16, 2019 at 2:08

This is a partial answer. I'm going to derive the equations satisfied by the center of desired ellipses. As of this moment,, solving the equations by anything non-numerical is out of reach. The main result is

The center lies on the intersection of two algebraic curves, one cubic and another quartic.

To avoid conflict with other uses of the variable $$a,b$$. I will use $$\alpha,\beta$$ to denote the semi-major/semi-minor axis for the desired ellipse.

main result - center lies on intersection of a cubic and a quartic curve.

First, let us determine the condition for the origin to be such a center. If $$\theta$$ is the angle between the semi-major axis with the $$x$$-axis.The equation of the ellipse will be

\begin{align} & {\small \frac{(x\cos\theta + y\sin\theta)^2}{\alpha^2} + \frac{(-x\sin\theta+y\cos\theta)^2}{\beta^2} = 1}\\ \iff & {\small \left(\frac{\cos^2\theta}{\alpha^2} + \frac{\sin^2\theta}{\beta^2}\right)x^2 + \left(\frac{\sin^2\theta}{\alpha^2} + \frac{\cos^2\theta}{\beta^2}\right)y^2 + \left(\frac{1}{\alpha^2}-\frac{1}{\beta^2}\right)\sin(2\theta)xy = 1}\\ \iff & {\small \frac12\left(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\right)(x^2 + y^2) + \frac12\left(\frac{1}{\alpha^2}-\frac{1}{\beta^2}\right)\left((x^2-y^2)\cos(2\theta) + 2xy\sin(2\theta)\right) = 1 } \end{align} Let $$\epsilon = \frac{\sqrt{\alpha^2-\beta^2}}{\alpha}$$ be the eccentricity of the ellipse and define $$\sigma$$ and $$\lambda$$ by

$$\frac{1}{\sigma^2} = \frac12\left(\frac1{\alpha^2} + \frac{1}{\beta^2}\right) \quad\text{ and }\quad \lambda = \frac{\alpha^2-\beta^2}{\alpha^2+\beta^2} = \frac{\epsilon^2}{2-\epsilon^2}$$

$$\sigma$$ and $$\lambda$$ will be an alternative measure of the size and eccentricity of the ellipse. In particular, for the conic above to be an ellipse, we need $$0 \le \epsilon < 1 \iff 0 \le \lambda < 1$$. In terms of them, above equation becomes

$$\frac{1}{\sigma^2}(x^2+y^2) - \frac{\lambda}{\sigma^2}\left[(x^2-y^2)\cos(2\theta)+2xy\sin(2\theta)\right] = 1$$

For any $$p = (x,y), q = (x', y') \in \mathbb{R}^2$$, define

• $$\tilde{p} = (-y,x)$$, the image of $$p$$ after a $$90^\circ$$ counterclockwise rotation.
• $$p \cdot q = xx' + y y'\;$$ be ordinary 2-d dot product.
• $$p \times q \stackrel{def}{=} \tilde{p}\cdot q = x y' - y x'\;$$ be the "cross product" for 2-d numbers.
• $$U(p) = \begin{bmatrix} x^2 + y^2\\ x^2 - y^2\\ 2xy\end{bmatrix}$$ and $$V(p) = U(\tilde{p}) = \begin{bmatrix} x^2 + y^2\\ -x^2 + y^2\\ -2xy\end{bmatrix}$$

In terms of them, the equation for 3 points $$A,B,C$$ to lie on the ellipse centered at origin is

$$\Lambda \cdot U(A) = \Lambda \cdot U(B) = \Lambda U(C) = 1\quad\text{ where }\quad \Lambda = \frac{1}{\sigma^2}\begin{bmatrix} 1\\ -\lambda\cos(2\theta)\\ -\lambda\sin(2\theta) \end{bmatrix}$$ With a little bit of vector algebra, we can solve above three equations to get

$$\Lambda = \frac{U(A)\times U(B) + U(B)\times U(C) + U(C)\times U(A)}{U(A)\cdot (U(B) \times U(C))}$$

It is not hard to verify the $$U(\cdot), V(\cdot)$$ satisfy following identities

• $$U(P)\cdot U(Q) = V(P)\cdot V(Q) = 2(P\cdot Q)^2$$.
• $$U(P)\cdot V(Q) = V(P)\cdot U(Q) = 2(P\times Q)^2$$.
• $$U(P)\times U(Q) = 2(P\times Q)\left[V\left(\frac{P+Q}{2}\right)-V\left(\frac{P-Q}{2}\right)\right]$$

Using these identities, the denominator simplifies to

\small \begin{align} & U(A)\cdot(U(B)\times U(C))\\ = & 2(B\times C)\,U(A)\times\left[V\left(\frac{B+C}{2}\right) - V\left(\frac{B-C}{2}\right)\right]\\ = & 2(B\times C)\left[ 2\left(A \times \frac{B+C}{2}\right)^2 - 2\left(A \times \frac{B-C}{2}\right)^2\right]\\ = & -4(B \times C)(C\times A)(A\times B) \end{align} Let $$\Delta$$ be the area of $$\triangle ABC$$ and $$u,v,w$$ be the barycentric coordinates of the origin with respect to $$\triangle ABC$$. We have

$$2\Delta u = B \times C,\quad 2\Delta v = C \times A\quad\text{ and }\quad 2\Delta w = A \times B$$

We find the denominator equals to $$-32\Delta^3 uvw$$.

Doing similar thing to the numerator and let $$D, E, F$$ be the midpoint of $$BC, CA$$ and $$AB$$, the numerator becomes

$$\small 4\Delta \left[ u (V(D) - V(E-F)) + v(V(E) - V(F-D)) + w(V(F)-V(D-E)) \right]$$ This is a little bit clumsy to write down. Let us use the notation $$\sum_{cyc}$$ to indicate a cyclic sum of over parameter set $$u \to v \to w,\quad A \to B \to C\quad\text{ and }\quad D \to E \to F$$ In this new notation, the equation for $$\Lambda$$ becomes

$$\Lambda = - \frac{1}{8\Delta^2 uvw} \sum_{cyc} u(V(D) - V(E-F))$$

Switch to other coordinate system where the center, let's call it $$Z$$, is no longer the origin, the equation for the center becomes

$$\Lambda = -\frac{1}{8\Delta^2 uvw}\sum_{cyc}u(V(Z-D) - V(E-F))\tag{*1}$$

To proceed further, we will do two things.

1. We will switch to a coordinate system where $$N$$, the nine-point center of $$\triangle ABC$$ is the origin. In this coordinate system, $$D,E,F$$ will be lying on a circle centered at origin $$N$$ with radius $$\frac{R}{2}$$ ($$R$$ is the circumradius of $$\triangle ABC$$).

2. We will identify the Euclidean plane with the complex plane, we will use the lower case letter to denote the complex number corresponds to a point. e.g. $$Z$$ becomes $$z$$ and $$D, E, F$$ become $$d, e, f$$.

After this, we can repress $$(*1)$$ as two equations

\begin{align} \Lambda_1 = \frac{1}{\sigma^2} &= -\frac{1}{8\Delta^2 uvw} \sum_{cyc} u(|z-d|^2 - |e - f|^2)\tag{*2a}\\ \Lambda_2 + i\Lambda_3 = -\frac{\lambda}{\sigma^2} e^{2i\theta} &= +\frac{1}{8\Delta^2 uvw} \sum_{cyc} u((z-d)^2 - (e - f)^2)\tag{*2b} \end{align} Let $$o = d + e + f$$ (the point corresponding to it is the circumcenter $$O$$ of $$\triangle ABC$$) and $$u',v',w'$$ the barycentric coordinate of $$Z$$ with respect to $$\triangle DEF$$. We have $$u = \frac{1-u'}{2},\cdots \implies \sum_{cyc} ud = \frac12\sum_{cyc}(1-u')d = \frac12(o-z)$$

Notice \begin{align} |z-d|^2 - |e-f|^2 &= |z-d|^2 + |e+f|^2 - 2|e|^2 - 2|f|^2\\ &= |z-d|^2+|o-d|^2 - 4|d|^2\\ &= |z|^2 -d(\overline{z+o}) - \bar{d}(z+o) + |o|^2 - 2|d|^2 \end{align}

Multiply by $$u$$ and take cyclic sum, we get

\begin{align} \sum_{cyc}u(|z-d|^2 - |e-f|^2) = & |z|^2 - \frac{(o - z)(\overline{z+o})}{2} - \frac{(\overline{o-z})(z+o)}{2} + |o|^2 - 2|d|^2\\ = & 2|z|^2 - \frac{R^2}{2}\end{align}

Equation $$(*2a)$$ becomes

$$\bbox[border:1px solid blue;padding: 1em;]{z\bar{z} - \frac{R^2}{4} + \frac{4\Delta^2 uvw}{\sigma^2} = 0} \tag{*3a}$$

Since $$u,v,w$$ are linear in $$(x,y) = (\Re z,\Im z)$$. This describe a cubic curve in the Euclidean plane and it is the locus of the center when the size parameter $$\sigma$$ is held fixed.

Let $$\Omega = \frac{def}{|d|^2} = \frac{4def}{R^2}$$. By a similar procedure, we have

\begin{align}(z-d)^2 - (e-f)^2 &= (z-d+e+f)(z-d-e-f) = (z-o+2e)(z-o+2f)\\ &= (z-o)^2+2(e+f)(z-o) + 4ef\\ &= z^2 - o^2 - 2d(z-o) + 4\Omega \bar{d} \end{align} Multiply by $$u$$ and take cyclic sum, we get

\begin{align}\sum_{cyc} u((z-d)^2 - (e-f)^2) &= z^2 - o^2 - (o-z)(z-o) + 2\Omega(\bar{o} - \bar{z})\\ &= 2\left(z(z-o) - \Omega(\bar{z} - \bar{o})\right)\end{align}

Equations $$(*2b)$$ becomes

$$z(z-o) - \Omega(\bar{z} - \bar{o}) = -\lambda e^{2i\theta}\frac{4\Delta^2 uvw}{\sigma^2}$$

Compare this with equation $$(*3a)$$, we obtain $$z(z-o) - \Omega(\bar{z} - \bar{o}) = \lambda e^{2i\theta} \left(z\bar{z} - \frac{R^2}{4}\right)$$ Taking absolute value and square, we obtain a quartic curve $$\bbox[border:1px solid blue;padding: 1em]{|z(z-o) - \Omega(\bar{z}-\bar{o})|^2 = \lambda^2 \left|z\bar{z} - \frac{R^2}{4}\right|^2}\tag{*3b}$$ This is the locus of the center when the eccentricity $$\epsilon$$ is help fixed.

observation I - locus for $$\sigma$$.

1. All locus of $$\sigma$$ passes through $$D,E,F$$ and the three foots of $$\triangle DEF$$.

2. When orthocenter $$H$$ of $$\triangle ABC$$ belongs to interior of $$\triangle DEF$$. $$\sigma$$ takes a local minimum at $$H$$. This should corresponds to a special circumellipse associated with $$\triangle ABC$$. I'm unable to figure out what that is. Anyone has any idea?

observation II - locus for $$\lambda$$.

1. When $$\epsilon = 0 \implies \lambda = 0$$, the locus reduces to a single point $$O$$. The circumcenter of $$\triangle ABC$$.

2. When $$\epsilon = 1 \implies \lambda = 1$$, the quartic terms in Eq. ($$*3b$$) cancel out. The corresponding curve reduces a union of $$3$$ lines through the midpoints. i.e. the lines $$DE$$, $$EF$$ and $$FD$$.

3. These $$3$$ lines split the plane into $$7$$ regions. When one varies $$\epsilon$$ (and hence $$\lambda$$) from $$0$$ to $$1$$. The locus sweep across the interior of $$4$$ out of the $$7$$ regions ( $$\triangle DEF$$ or the $$3$$ cone with apex at $$D, E, F$$). This means in order for a point to be center for an ellipse, it need to be either $$D, E, F$$ or belongs to the interior of above $$4$$ regions.

4. Infinitely many loci of $$\lambda$$ passing through $$D, E, F$$. Combine with the observation $$I_1$$, $$D, E, F$$ are centers for infinitely many circumellipses of $$A,B,C$$.

To be continued???

• Impressive effort, but please have a look at my solution. There is a shorter way to a univariate polynomial equation.
– user65203
Jun 19, 2019 at 7:28
• @YvesDaoust I've seen your answer but I'm more interested in the geometry side of the problem. I want to understand what configuration is possible and how to classify the solutions (either geometrically or topologically). Jun 19, 2019 at 8:42

WLOG, one of the given points is the origin (otherwise, apply a translation).

Let us hypothesize the angle $$\theta$$. We counter-rotate the three points so that the axis $$a$$ becomes horizontal. Then we apply an anisotropic dilation of ratio $$r:=\dfrac ab$$ on the $$y$$ coordinate. This way, the ellipse becomes a circle of radius $$a$$, through the origin. The transformation is

$$\begin{cases}p_i=x_i\cos\theta+y_i\sin\theta,\\q_i=r\left(-x_i\sin\theta+y_i\cos\theta\right).\end{cases}$$

The implicit equation of a circle through the origin is

$$\begin{vmatrix}p^2+q^2&p&q\\p_1^2+q_1^2&p_1&q_1\\p_2^2+q_2^2&p_2&q_2\end{vmatrix}=0$$ or

$$\Delta(p^2+q^2)+\Delta_p p+\Delta_q q=0$$ where the deltas are minors.

By completing the square,

$$\left(p-\frac{\Delta_p}{2\Delta}\right)^2+\left(q-\frac{\Delta_q}{2\Delta}\right)^2=\frac{\Delta_p^2+\Delta_q^2}{4\Delta^2}$$ and in this expression the RHS is the squared radius.

By equating to $$a^2$$,

$$(q_1(p_2^2+q_2^2)-q_2(p_1^2+q_1^2))^2+(p_1(p_2^2+q_2^2)-p_2(p_1^2+q_1^2))^2=4a^2(p_1q_2-p_2q_1)^2.$$

Unfortunately, this expands to a big sextic polynomial in $$\cos\theta,\sin\theta$$ and I don't foresee a nice simplification.

Note that you can rationalize the last equation by

$$\begin{cases}\cos\theta=\dfrac{t^2-1}{t^2+1},\\\sin\theta=\dfrac{2t}{t^2+1}\end{cases}$$

and you will end up with a univariate polynomial of degree $$12$$. But by symmetry, if $$\theta$$ is a solution, so is $$\theta+\pi$$ and the $$t$$ solutions will come in pairs $$t\,t'=-1$$.

Let the three points be $$A,B,C$$. The idea of this solution is to transform $$\triangle ABC$$ into an equilateral triangle with vertices $$(0,0), (1,0), (\dfrac{1}{2}, \dfrac{\sqrt{3}}{2} )$$

There are three possible arrangements for the vertices in terms of the pre-images of the equilateral triangle vertices:

$$A , B , C$$

$$A , C , B$$

$$B , C , A$$

The affine transformation is

$$f(P) = T (p - V_1)$$

where $$V_1$$ is the first vertex.

Take for example the first arrangement, then

$$T \begin{pmatrix} B- A && C - A \end{pmatrix} = \begin{pmatrix} 1 && \dfrac{1}{2} \\ 0 && \dfrac{\sqrt{3}}{2} \end{pmatrix}$$

From which, we can calculate the transformation matrix $$T$$.

Now, we'll pass an ellipse in standard orientation through the three vertices of the equilateral triangle, let the center of the ellipse be $$C = (1/2 , c )$$, then the equation of ellipse

$$\dfrac{(x - (1/2))^2}{ a^2 } + \dfrac{(y - c)^2} {b^2} = 1$$

The points through which this ellipse passes are $$(0,0), (1,0), (\dfrac{1}{2}, \dfrac{\sqrt{3}}{2} )$$

Substituting the first is the same as substituting the second, they both give us

$$\dfrac{1}{4 a^2} + \dfrac{c^2}{ b^2} = 1$$

Substituting the third point,

$$\dfrac{ (\sqrt(3)/2 - c ) ^2}{ b^2 }= 1$$

So that $$b = sqrt(3)/2 - c$$

Now define the matrix

$$Q = \begin{bmatrix} \dfrac{1}{a^2} && 0 \\ 0 && \dfrac{1}{b^2} \end{bmatrix}$$

The pre-transformation Q is obtained as follows:

The position vectors are related by $$r = T (r' - A)$$ where $$r'$$ is the position vector of the original points, and $$r$$ is the position vector after applying the transformation.

Now, we have the ellipse that passes through the three vertices of the equilateral triangle is given by

$$(r - C)^T Q (r - C) = 1$$

where $$C = (1/2, c )$$

Substitute $$r$$

$$(T (r' - A) - C )^T Q (T (r' - A) - C) = 1$$

And this simplifies to

$$(r' - A - T^{-1} C ) ^T T^T Q T ( r' - A - T^{-1} C ) = 1$$

so our $$Q' = T^T Q T$$

Expanding,

$$Q' = \begin{bmatrix} (1/a^2) t_1^2 + (1/b^2) t_3^2 && (1/a^2) t_1 t_2 +(1/b^2) t_3 t_4 \\(1/a^2) t_1 t_2 +(1/b^2) t_3 t_4 && (1/a^2) t_2^2 + (1/b^2) t_4^2 \end{bmatrix}$$

where $$T = \begin{bmatrix} t_1 && t_2 \\ t_3 && t_4 \end{bmatrix}$$

we want the eigenvalues of $$Q'$$ to be $$1/a_e^2$$ and $$1/b_e^2$$ where $$a_e, b_e$$ are the given semi-axes lengths of the ellipse.

so using the trace and the determinant, we obtain

$$\dfrac{1}{a^2} (t_1^2 + t_2^2) + \dfrac{1}{b^2} (t_3^2 + t_4^2 ) = \dfrac{1}{a_e^2} + \dfrac{1}{b_e^2} \hspace{30pt}(1)$$

and for the determinant expression, it simplifies to

$$\dfrac{1}{a b} | t_1 t_4 - t_2 t_3 | = \dfrac{1}{a_e b_e} \hspace{30pt} (2)$$

Now recall that

$$\dfrac{1}{4 a^2} + \dfrac{c^2}{ b^2} = 1$$

and that

$$c = \dfrac{\sqrt{3}}{2} - b$$

substituting this last equation in the previous one, gives

$$\dfrac{1}{4 a^2} + \dfrac{3}{4 b^2} - \dfrac{\sqrt{3}}{b} = 0 \hspace{30pt}(3)$$

This system of equations $$(1), (2), (3)$$ in the two unknowns $$a , b$$ is overdetermined. So the strategy to solve it is to solve $$(1), (2)$$ and then check that $$(3)$$ is satisfied. Once all three equations are satisified then we have our ellipse, and we can compute the center and the $$Q'$$ matrix from which by diagonalizing we can compute $$\theta$$.