How to get 5 points of an ellipse which is internaly tangent to two congruent intersecting circles.

Question

Let two circles $C$ and $C’$ intersecting at points $A$, $B$. I would like to construct an ellipse passing through $A$ and $B$ using the $5$ points construction of GeoGebra (foci unknown). The problem is that the ellipse must have two points of tangency ($E$ with $C$ and $F$ with $C’$). Do you know a method that uses inversion ? Is there a way to impose this tangency ? Many thanks to any hint/help.

Answer 1

I'm assuming the circles to be congruent, otherwise the problem has no solution, in general. Point $E$ can then be chosen at will, while $F$ is the reflection of $E$ about the midpoint $S$ of $AB$, which is also the center of the ellipse.

Draw the line $t$ tangent to the circle at $E$, then draw through $S$ a line $r$ parallel to $t$ and call $M$, $N$ the (unknown) intersections between $r$ and the ellipse. $EF$ and $MN$ are conjugate diameters.

Draw a point $P$ on $EF$ such that $AP\parallel r$. By the properties of conjugate diameters we have then: $$ {AP^2\over SM^2}+{SP^2\over SF^2}=1. $$ From that you can compute $SM$: $$ SM={AP\over\sqrt{1-SP^2/SF^2}} $$ and construct point $M$ on line $r$.

Answer 2

Assuming the two circles are congruent, and that the tangents at $E$ and $F$ are parallel, then from symmetry, we know that the center of the ellipse is at the midpoint of $A$ and $B$. Let this point be $M_{AB}$, i.e.

$ M_{AB} = \dfrac{1}{2} (A + B ) = (x_1, y_1) $

In addition to the center $M_{AB}$ we know that point $A$ is on the ellipse. And we also have point $E$, and finally we know the slope of the tangent at point $E$. The equation of the ellipse satisfying these conditions can be derived from the equation of an ellipse centered at $M_{AB}$.

$ (r - M_{AB})^T Q (r - M_{AB} ) = 1 $

where $ r = [x, y]^T $ and $Q = \begin{bmatrix} a && b \\ b && c \end{bmatrix}$

Expanding,

$ a x^2 + 2 b x y + c y^2 - 2 a x x_1 - 2 b ( x y_1 + y x_1) -2 c y y_1 + a x_1^2 + 2 b x_1 y_1 + c y_1^2 -1 = 0 $

Which is nice and linear in the $3$ unknowns: $a,b$, and $c$.

Using the coordinates of $A$ and $E$ gives two independent equations in the three unknowns.

Next, differentiate the above equation implicitly with respect to x, this will give

$ 2 a x + 2 b (x y' + y ) + 2 c y y' - 2 a x_1 - 2 b (y_1 + y' x_1) -2 c y' y_1 = 0 $

Substitute point $E$ and the slope at point $E$ into this equation, which provides the third independent equation.

Solve the system of $3$ equations in the $3$ unknowns, and this gives the values of $a,b$ and $c$.

Having obtained $a,b,c$, you now have the equation of the ellipse, which you can feed to Geogebra to obtain a plot.