How to draw ellipse and circle tangent to each other?

The circle $c$ is given as are the points $A$ and $B$, which are ellipse's foci. Now I need to construct the ellipse that is tangent to the circle $c$ such that the points $A$ and $B$ are its foci.

Actually I'm interested in the point of tangency, so it would be enough to construct it. But please note that I don't want any algebraic calculation: all I need is a purely geometric solution.

I've searched over the Internet and I found that at the point of tangency the ray $OX$, where $O$ is circle's center and $X$ is the point of tangency, should bisect the angle $AXB$. This may be helpful, but please note that I don't have any proof for this; but I've run computer examples and it was true for all of them.

Solution 1:

This is an interesting problem if one insists on a solution constructable using purely geometric operations.

For simplicity in describing the solution, we will choose a coordinate system such that the two foci $A, B$ are $(1,0)$ and $(0,1)$ respectively. We will also assume the circle $\mathscr{C}$ we are dealing with is centered at $C = (\alpha,\beta)$ in the first quadrant and the radius of $\mathscr{C}$ is $r$.

It is known that the family of ellipses and hyperbolas having $A$ and $B$ as foci are given by:

$$\frac{x^2}{1+\lambda} + \frac{y^2}{\lambda} = 1\quad\text{ and is } \begin{cases} \text{ an ellipse, }&\text{ if } \lambda > 0\\ \text{ a hyperbola, } &\text{ if } 0 > \lambda > -1 \end{cases}\tag{*0}$$ For a point $p = (x,y)$ on this ellipse/hyperbola, the normal is in the direction $(\frac{x}{1+\lambda}, \frac{y}{\lambda})$. i.e. the normal line passing through $(x,y)$ is given by:

$$\left\{\;\left( x ( 1 + \frac{t}{1+\lambda} ),\,y ( 1 + \frac{t}{\lambda} ) \right) : t \in \mathbb{R}\;\right\}$$

If this ellipse/hyperbola is tangent to any circle centered at $C$ at point $p$, the above normal line with pass through $C = (\alpha,\beta)$ and one can show that $(x,y)$ lies on a cubic curve $\mathscr{P}$:

$$(\alpha y - \beta x )\left(x(x-\alpha) + y(y-\beta)\right) - (x-\alpha)(y-\beta) = 0\tag{*1}$$

This is the red curve depicted in above picture.

For $r$ not too large, $\mathscr{C}$ and $\mathscr{P}$ in general will intersect at 4 points. Two of them are "tangent" points to two ellipses and the other two are tangent points to two hyperbolas. All these four ellipses/hyperbolas are having $A, B$ as their foci. For these four points, one can simplify $(*1)$ and shows that they lie on a conic $\mathscr{Q}$: $$(\alpha y - \beta x)(\alpha x + \beta y - \gamma) - (x-\alpha)(y-\beta) = 0\tag{*2}$$ where $\gamma = \alpha^2 + \beta^2 - r^2$. This is the cyan curve containing $C, F, G, J, K, L$ in above picture.

Some of these points are relatively easy to construct.

• $C$ - of course we know where $C$ is. In fact, one can show that $\mathscr{Q}$ is tangent to the line $OC$ at point $C$. Counting multiplicity, this gives "two" points to determine $\mathscr{Q}$.
• If one draws a circle using $OC$ as diameter, it will intersect $\mathscr{C}$ at two points $D$ and $E$. The line $DE$ in turn intersects the horizontal line containing $C$ at $F$ and the vertical line containing $C$ at $G$.

A conic is determined by any five points on it. If for whatever means we get an extra point on $\mathscr{Q}$, then we can construct $\mathscr{Q}$ and its intersections with $\mathscr{C}$, e.g. the point $L$ in the above picture. These are the tangent points we want.

One can show that the conic $\mathscr{Q}$ intersects the $x$-axis at $J = (\mu,0)$ and $K = (1/\mu,0)$ where $\mu$ is a root of the quadratic equation: $$\alpha x^2 - (\gamma +1) x + \alpha = 0\tag{*3}$$ If one doesn't mind a little bit of algebra, we are done. Otherwise, we can convert this formula to a geometric construction of $J$ and $K$ in 3 steps:

1. Construct an auxiliary hyperbola $xy + 1 = 0$ (the olive green hyperbola in above picture). The foci of this hyperbola are $(\sqrt{2},-\sqrt{2})$ and $(-\sqrt{2},\sqrt{2} )$ and it contains the point $(1,-1)$. All of them can be constructed using compass and ruler start from $O$ and $A$.
2. Construct an auxiliary horizontal line $\beta y + 1 = 0$ (the blue horizontal line containing $I$ in above picture). One first constructs a line segment joining $Y = (0,\beta)$ to $B = (-1,0)$. One then constructs another line segment parallel to it start at $Y' = (0,1)$. This new segment will intersect the $x$-axis at $B' = (-\frac{1}{\beta},0)$. Another quarter circle will bring us to the point $I = (0,-\frac{1}{\beta})$ and the horizontal line at $I$ is the one we need.
3. Let $H$ be the intersection of the line $DE$ with the blue line in previous step. Project it onto $x$-axis as $H_1$. Now construct a ray start from $H_1$ at $-135^{\circ}$ with respect to the $x$-axis. This ray will intersect the "olive green" hyperbola at $H_2$ and $H_3$. If one projects $H_2$ and $H_3$ back to the $x$-axis, we get $K$ and $J$ respectively.

Update Since Stefan asks, here are some details how to derive the various equations.

If the normal passes through $C = (\alpha,\beta)$, we have a $t$ such that:

$$x (1 + \frac{t}{1+\lambda}) = \alpha \quad\text{ and }\quad y (1+ \frac{t}{\lambda}) = \beta$$ Eliminating $t$ give us: $$\begin{align}(1+\lambda)(\frac{\alpha}{x}-1) = \lambda (\frac{\beta}{y} - 1 ) \iff & (1+\lambda)(x-\alpha)y - \lambda(y-\beta)\alpha = 0\\ \iff & (x-\alpha)y - \lambda( \alpha y - \beta x ) = 0 \end{align}$$ This gives us $$ \lambda = \frac{(x - \alpha)y}{\alpha y - \beta x} \quad\text{ and }\quad 1 + \lambda = \frac{(y - \beta)x}{\alpha y - \beta x} $$ Substitute this back into $(*0)$, we get $(*1)$: $$\begin{align} & (\alpha y - \beta x)\left(\frac{x}{y-\beta} + \frac{y}{x-\alpha}\right) = 1\\ \iff & (\alpha y - \beta x)\left(x(x-\alpha)+y(y-\beta)\right) - (x-\alpha)(y-\beta) = 0 \end{align}$$

If $(x,y)$ also lies on $\mathscr{C}$, the $2^{nd}$ factor in above expression can be simplified as:

$$\begin{align} x(x-\alpha)+y(y-\beta) = & (x-\alpha)^2 + (y-\beta)^2 + \alpha (x-\alpha)+\beta(y-\beta)\\ = & \alpha x + \beta y - (\alpha^2 + \beta^2 - r^2)\\ = & \alpha x + \beta y - \gamma \end{align}$$

Substituting this into $(*1)$ gives us the equation of $\mathscr{Q}$ in $(*2)$:

$$(\alpha y - \beta x)(\alpha x + \beta y - \gamma) - (x-\alpha)(y-\beta) = 0$$

Now $(*3)$ is really the equation of intersecting $\mathscr{Q}$ with the $x$-axis. We just set $y = 0$ in $(*2)$, eliminate the common factor $\beta$ and we are done.