The area enclosed by the locus of point C Triangle ABC is such that AB = 4, BC = 2, and AC = 3. If vertex A is confined to the x-axis and vertex B is confined to the y-axis, what is the area of the region enclosed by the locus of all points point C could possibly be?
 A: You have some sort of a "do-nothing machine"
The midpoint of $AB$ traces a circle.
Lets start with $B = (0,0), A = (4,0)$  Where is $C$
Area  $= \sqrt {(\frac 92)(\frac 52)(\frac 32)(\frac 12)} = \frac {\sqrt {135}}{4}$ 
$h = \frac {\sqrt {135}}{8}$
And from Pythagoras theorem we get that the altitude splits the base into segments $\frac {11}{8},\frac {21}{8}$ 
Note there are 2 ways we can orient the triangle at this time.
$(\frac {11}{8},\frac  {\sqrt {135}}{4})\\
(\frac {11}{8},-\frac  {\sqrt {135}}{4})\\
$
Lets pick $C = (\frac {11}{8},\frac  {\sqrt {135}}{4})$ for now.
$A = (4\cos\theta, 0)\\
B = (0, 4\sin\theta)\\
C = (\frac {11}{8}\cos\theta + h\sin\theta, \frac {21}{8}\sin\theta + h\cos\theta)$
$C = (2\cos(\theta - \arctan\frac {\sqrt{135}}{11}), 3\sin (\theta + \arctan \frac{\sqrt {135}}{21}))$
That will trace out something that is roughly an ellipse.  I don't think it will be exactly an ellipse.
And swapping the different starting position of $C$ will change the sign on the angle of orientation.
