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?)