Orthogonal tangents to an ellipse This is the problem I found back in the first year in the university.
Suppose we have a non-degenerate (i.e. not a point and not an empty set) ellipse $E\subset \Bbb R^2$. Now define a set $D$ by a property $$(x,y)\in D\iff \text{tangent lines to }E \text{ through} (x,y)\text{ are orthogonal}.$$
We are to completely describe $D$ in terms of characteristics of $E$.
The answer is (don't look if you don't want spoilers) 

 that $D $ is a circle with the same center as $E$. Its radius is $\sqrt{a^2+b^2}$ where $a$ and $b$ are short and long radii of the ellipse.

The proof I found back then used coordinates and required a lot of boring algebraic maniplations. I wonder if there exist an elegant (definition of elegance is up to you) solution to this problem.
 A: 
I will provide a geometric proof.
Consider an Ellipse of foci $F$ and $F'$.
Let $M$ be a point outside the ellipse. The tangents from $M$ touch the Ellipse at $A$ and $A'$. Let $E$ be the symmetric of $F$ with respect to $MA$ and define $E'$ similarly.
${\bf Step 1.}$ The points $F'$, $A$ and $E$ are aligned. Indeed,
by the optical property of the ellipse $\angle MAF'=\angle FAX=\angle XAE$. Similarly, the $F$, $A'$ and $F'$ are also aligned.
${\bf Step 2.}$   $\triangle  FE'M$ and $\triangle F'EM$ are congruent. Because,
$EF'=EA+AF'=FA+AF'=2a$ and similarly, $FE'=2a$. Moreover, $ME=MF$ and $ME'=MF'$.
${\bf Step 3.}$ $\angle AMA'=\angle F'ME$. Indeed, from the previous step we conclude that
$$\angle XME=\frac{1}{2}\angle EMF=\frac{1}{2}\angle E'MF'=\angle YMF'.$$
${\bf Step 4.}$ It follows that $MA\bot MA'$ if and only if $\angle EMF'=\frac{\pi}{2}$,
and (since $EM=FM$,) this equivalent to 
$$FM^2+F'M^2=F'E^2=4a^2\tag{1}$$
But using the parallelogram identity we know that
$$
FM^2+F'M^2=2OM^2+2OF^2=2OM^2+2c^2
$$
Thus, $(1)$ is equivalent to $OM^2=a^2+b^2$, which is the desired conclusion.
A: For the sake of interest, you're describing something more commonly known as a director circle or the Fermat–Apollonius circle.
A fairly nice elementary proof involves the use of the discriminant and Vieta's Formulas. However, this very well may be the 'boring' proof that you mentioned ='(
Let $L:y=mx+c$ be a tangent to the canonical ellipse $E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$. Substitute $L$ into $E$ and you will have a quadratic in $x$. Since $L$ is tangential to $E$, we can say that the discriminant of this quadratic is zero, where you will arrive at the result $c^2=a^2m^2+b^2$ (after a bit of algebra and cleaning up!).
So now we can re-write $L$ as $y=mx\pm \sqrt{a^2m^2+b^2}$.
$y-mx=\pm \sqrt{a^2m^2+b^2} \\ (y-mx)^2=a^2m^2+b^2 \\ y^2 - 2xym+m^2x^2=a^2m^2+b^2 \\ m^2(a^2-x^2)+2xym+(b^2-y^2)=0$
Now suppose that this $L$ passes through some arbitrary external point $T(X,Y)$, then our quadratic in $m$ as above is $m^2(a^2-X^2)+2XYm+(b^2-Y^2)=0$
Geometrically, the solutions of this quadratic are the gradients of the tangents from $E$ that pass through $T$. 
However, we want these gradients to be perpendicular to each other. In other words, we want the product of the roots of this quadratic (in $m$) to be equal to $-1$. This is where Vieta's formulas kick in.
Using Vieta's formulas, we have $\dfrac{b^2-Y^2}{a^2-X^2}=-1 \Rightarrow b^2-Y^2=X^2-a^2 \Rightarrow X^2+Y^2=a^2+b^2$
This means that the locus of all orthogonal tangents is the circle (director circle) $x^2+y^2=a^2+b^2$, which has radius $\sqrt{a^2+b^2}$ as required.
(can somebody please fix up my TeX so the chunk equations is centred and aligned?) 
