Proving that the Concentric Circle Method constructs an Ellipse

Question

I'm in my second year of an engineering drawing course, and today our professor taught us how to construct an ellipse by a method called the Concentric Circle Method (steps shown below).

My question is, if one was given this method unknowing that the figure obtained was an ellipse, how could we prove that it was one?

I had a hunch that this might be a proof derived by using $x^2+y^2=a^2$ and $x^2+y^2=b^2$ as two base equations, and using them to prove the Standard Form of the ellipse: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

I'm not sure though. I would really appreciate any help or opinions anyone has on this.

Answer 1

Note: This construction is known as de La Hire's point construction.

Here we show the construction presented by OP produces points on the ellipse $E$.

We consider two concentric circles with center $(0,0)$. One of them is an incircle $C_1$ with radius $r$ of an ellipse $E$ and the other is an excircle $C_2$ with radius $R$ of the ellipse $E$. The ellipse $E$ has semi-minor axis $r$ and semi-major axis $R$ where $0<r<R$. \begin{align*} C_1:& x^2+y^2=r^2\tag{1}\\ C_2:& x^2+y^2=R^2\tag{2}\\ E:& \frac{x^2}{R^2}+\frac{y^2}{r^2}=1\tag{3} \end{align*}

We draw a line from the center $(0,0)$ with angle $t \in[0,2\pi)$.

• This line intersects the small circle at the point $\color{blue}{P_1=(r\cos t, r\sin t)}$. The point $P_1$ is an element of $C_1$, since \begin{align*} (r\cos t)^2+(r\sin t)^2=r^2\left(\cos^2t + \sin^2t\right)=r^2 \end{align*} according to (1).

• The line intersects the large circle at the point $\color{blue}{P_2=(R\cos t, R\sin t)}$. The point $P_2$ is an element of $C_2$, since \begin{align*} (R\cos t)^2+(R\sin t)^2=R^2\left(\cos^2t + \sin^2t\right)=R^2 \end{align*} according to (2).

                                      

The points constructed in OPs step 3 correspond here to the construction of the point $P_E$ \begin{align*} \color{blue}{P_E=(R\cos t, r\sin t)} \end{align*} Putting $P_E$ in (3) we obtain \begin{align*} \color{blue}{\frac{R^2\cos^2 t}{R^2}+\frac{r^2\sin^2 t}{r^2}=\cos^2 t + \sin^2 t =1} \end{align*} which verifies the construction.

Answer 2

Some geometrical intuition: from step 3 to 4, you are basically taking each point in the large circle and dividing its $y$ coordinate by the ratio between the radii of the circles. Therefore, with the procedure shown in the figures, you go from the points that satisfy $$x^2 + y^2 = R^2$$ to those that satisfy $$x^2 + \left(\frac{R}{r}y'\right)^2 = R^2$$ This is nothing but an ellipse with a semi-minor axis equal to $r$

Answer 3

First, for convenience, let's assume the center of your ellipse is at $(0,0)$ in a Cartesian coordinate plane and that the point $B$ is on the positive $x$-axis in that plane. That is, the major axis of the ellipse is on the $x$-axis and the minor axis of the ellipse is on the $y$-axis.

Further, suppose $a$ is the semi-major axis of the ellipse (so $B$ is at $(a,0)$) and $b$ is the semi-minor axis (so $C$ is at $(0,b)$).

The point $B$ will be on the ellipse, of course, because it is one end of the major axis of the ellipse.

Now let's look at the next point in counterclockwise order, which is related to the radial line that intersects the outer circle just above $B$, at $22.5^\circ$ above the positive $x$-axis. This radial line intersects the inner circle at $(b\cos 22.5^\circ, b\sin 22.5^\circ)$ and intersects the outer circle at $(a\cos 22.5^\circ, a\sin 22.5^\circ)$.

From the intersection with the outer circle, we drop a vertical line which has $x$-coordinate $a\cos 22.5^\circ$. From the intersection with the inner circle, we extend a horizontal line which has $y$-coordinate $b\sin 22.5^\circ$.

Therefore the point where those two lines intersect has coordinates $$(x,y) = (a\cos 22.5^\circ, b\sin 22.5^\circ). $$

Similarly with the other points that we plot via this construction; they are either already at the ends of axes of the ellipse or they are found by intersecting lines that intersect at a point $$(x,y) = (a\cos \theta, b\sin \theta) \tag1$$ where $\theta$ is one of the angles of the radial lines, such as $45^\circ$, $67.5^\circ$, $112.5^\circ$, and so forth. (Actually even the ends of the axes can be written like this by setting $\theta$ to $0^\circ$, $90^\circ$, $180^\circ$, or $270^\circ$.)

So every point that we plot in this way has coordinates specified by Equation $(1)$ for some angle $\theta$. These are precisely the points of an ellipse, because if we set $x$ and $y$ according to Equation $(1)$, then

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2} = \frac{a^2\cos^2\theta}{a^2}+\frac{b^2\sin^2\theta}{b^2} = \cos^2\theta + \sin^2\theta = 1. $$