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.

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    $\begingroup$ What you say in the ladt paragraph is correct and due to the reflectivity property of the ellipse: A light ray coming out of one focus reflects off the ellipse back to the other focus. So $AX$ and $BX$ make equal angles with the normal to the ellipse at $X$. $\endgroup$ – Ted Shifrin Aug 11 '13 at 3:05
  • $\begingroup$ I get it, but I don't know how this fact can help me with the construction $\endgroup$ – Stefan4024 Aug 11 '13 at 3:15
  • $\begingroup$ Do we assume the center of the circle called $c$ is the midpoint of the segment joining the foci $A,B$? $\endgroup$ – coffeemath Aug 11 '13 at 12:34
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    $\begingroup$ No, foci $A$ and $B$ must be outside of the circle, so they can be completely on one side of the circle and the segment that joins them will not go through the circle $\endgroup$ – Stefan4024 Aug 11 '13 at 12:50

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

Circle kissing Ellipse v2

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.

  • $\begingroup$ Great work mate, it's very clearly explained, although I must admit I thought I would understand it at the beggining. Just one more question. What if line $AB$ isn't parallel to the x-axis, will we get $K$ and $L$ using the same method? I mean would all those parametars be same? $\endgroup$ – Stefan4024 Aug 14 '13 at 13:42
  • $\begingroup$ @Stefan4024 As long as you interpret horizontal as something parallel to the major axis, vertical as something parallel to the minor axis, the construction is essentially independent of choice of direction. $\endgroup$ – achille hui Aug 14 '13 at 13:51
  • $\begingroup$ Let's say segment $AB$ isn't parallel to any axis will all this parametar be same? One question how you prove that those 4 points lie on a conic? And how the step 3 give point $K$ and $L$? Why? $\endgroup$ – Stefan4024 Aug 14 '13 at 15:19
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    $\begingroup$ If $AB$ isn't parallel to any axis, the parameter $\alpha,\beta$ will be different. In fact, the cubics $\mathscr{P}$ and the conics $\mathscr{Q}$ no longer have the simple form I gave here. However, the constructed points maintain same set of geometric relations among each other. We don't need to prove $F, G, K, L$ lies on a conic. We already have a conic in $(*2)$ and we intersect it with various lines and "find" the points. In step 3, we have a quadratic equation to solve. The steps are created to mimic the process of setting up and solving the quadratic equation in algebra! $\endgroup$ – achille hui Aug 14 '13 at 16:18
  • $\begingroup$ 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 $\mathscr{P}$: $$(\alpha y - \beta x )\left(x(x-\alpha) + y(y-\beta)\right) - (x-\alpha)(y-\beta) = 0\tag{*1}$$ How an you prove that? And how do you simplify it later? And how do you obtained that quadratic solution? $\endgroup$ – Stefan4024 Aug 14 '13 at 17:42

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