Area of circle

If we desire to circumscribe a right triangle with side lengths $a, b$ , then the hypothenuse $\sqrt{a^2 + b^2}$ is the diameter of this circumcircle (as the inscribed right angle is half of a semicircle). This poses the following LaGrange Multiplier problem:

Minimize $C(a,b) = \frac{\pi}{4}(a^2 + b^2)$ , subject to $T(a,b) = \frac{1}{2}ab = S$ .

Taking $grad (C) = \lambda \cdot grad (T)$ , we obtain:

$\frac{\pi a}{2} = \lambda \frac{b}{2}$

$\frac{\pi b}{2} = \lambda \frac{a}{2}$

or $\frac{b}{a} = \frac{a}{b} \Rightarrow a = b$ . Calculating $T(a,a) = \frac{a^2}{2} = S \Rightarrow a = b = \sqrt{2S}.$

The Hessian matrix of C easily computes to $\frac{\pi}{2} \cdot I_{2x2}$ (where $I_{2x2}$ is the 2x2 identity matrix) and is positive-definite for all $a,b \Rightarrow a = b = \sqrt{2S}$ is a global minimum point. This ultimately yields the minimum circular area:

$C(\sqrt{2S}, \sqrt{2S}) = \frac{\pi}{4} \cdot 2(\sqrt{2S})^2 = \boxed{\pi S}.$