Let $O$ be a circle of radius $r$ lying in the first quadrant of the Cartesian plane, tangent to the $x$-axis at the point $(r, 0)$ and tangent to the $y$-axis at the point $(0, r)$. Let $T$ be a right triangle whose legs are segments of the axes, with inscribed circle $O$. What is the ratio of the legs of the triangle when the area of $T$ is minimized?
It seems obvious that the answer is $1$ (the right triangle is isosceles), but I'm having a time showing this.
We can draw a diameter through the circle parallel to the $x$-axis. Let $p$ be the point of tangency between the circle and a hypotenuse, and let $\theta$ be the angle between the diameter and the radius meeting $p$. Since the radius meeting $p$ will be perpendicular to the hypotenuse, the hypotenuse will have slope $-\cot\theta$. Letting $(x, 0)$ and $(0, y)$ be the axis intercepts of the hypotenuse, and $[(r+\cos\theta), (r+\sin\theta)]$ the point $p$ of tangency, we have
$$x=r+\cos\theta+r\tan\theta+\sin\theta\tan\theta \\ y=r+\sin\theta+r\cot\theta+\cos\theta\cot\theta.$$
We multiple those (yikes), differentiate, and solve. I did that, thinking that eventually I'd get a simpler expression. No such luck.
Is there a much simpler way to do this that I'm not seeing?