Line And Locus

All right triangles can be inscribed inside a circle where the hypotenuse is the diameter and the vertex of the $90^{\circ}$ angle lies somewhere on the circle's circumference.

Since $P$ is the $90^{\circ}$ angle, $AB$ is the hypotenuse. This means the locus of points is a circle with diameter $l$ , so the area is $\pi \left ( \dfrac{l}{2} \right ) ^2 = \large \color{#20A900}{\boxed{ \dfrac{\pi l^2}{4}}}$ .

Let's try a vector approach to this locus problem. Define the points $A(0,0), B(l,0), P(x_{0}, y_{0})$ in the $xy-$ plane. Taking the vectors:

$PA = -x_{0} \hat{i} - y_{0} \hat{j}$

$PB = (l-x_{0}) \hat{i} - y_{0} \hat{j}$

the angle $\angle{APB}$ will be a right angle iff $PA \cdot PB = 0,$ or:

$(-x_{0})(l-x_{0}) + (-y_{0})^{2} = 0$ ;

or $x^{2}_{0} - lx_{0} + y^{2}_{0} = 0;$

or $x^{2}_{0} - lx_{0} + \frac{l^2}{4} + y^{2}_{0} = \frac{l^2}{4};$

or $(x_{0} - l/2)^2 + y^{2}_{0} = (l/2)^2$

which is just a circle with radius $\frac{l}{2}$ and area of $\boxed{\frac{\pi l^2}{4}}.$