Largest rectangular area formed by origin and a point on a circle Lets say the circle has an equation of $(x-2)^2+(y-2)^2=4$.


How can I algebraically derive the largest possible rectangular area formed by the two corners (0, 0) and (x, y) where x and y are points on the circle?
Here are a couple of (educated) intuitions to explain why the answer must occur at a point where $x = y.$



Consider the area of a rectangle in the first quadrant with opposite vertices at $(0,0)$ and $(x,y)$ and edges parallel to the axes.
If the area is $1$ (for example), the point $(x,y)$ lies somewhere on the branch of the hyperbola $xy = 1$ in the first quadrant.
This curve is symmetric around the line $y = x,$ and lies above the line
$y = 2 - x$ except at the point of tangency, $(1,1).$
This is the level curve for area equal to $1.$

Every other level curve for the area of the rectangle has a similar shape:
a branch of a hyperbola in the first quadrant, symmetric around the line
$xy = k$ for some constant $k,$ tangent at a point of the form $(x,x)$
to a line $y = b - x$ for some constant $b.$

So find the points where the circle intersects the line $y = x,$ and
pick the one farthest from the origin; one of the level curves is
tangent to a line with slope $-1$ through that point, which is also
tangent to the circle.
Any other level curve that intersects the circle will lie below this one, and therefore any other point on the circle leads to a rectangle of smaller area.



Another approach: 
Suppose the a rectangle has opposite vertices at $(0,0)$ and $(x,y),$
where $(x,y)$ is in the first quadrant.
If we do not make the restriction that the edges of the rectangle are parallel to the axes, then the largest possible area of the rectangle is the area of the square with diagonal from  $(0,0)$ to $(x,y).$

The area of that square is larger when $(x,y)$ is farther from the origin.
The area is maximized when $(x,y)$ is as far from the origin as it can get,
which is on a circle centered at the origin such that at least one point of the given circle is on the larger circle but all the other points of the given circle lie inside the larger circle.
The large circle is in fact the circle tangent to the given circle
at the farther point where the given circle intersects the line $y = x.$

So we choose the square with diagonal ending at that point.
No other square with diagonal ending on the circle is larger,
and for each of those squares, the rectangle with the same diagonal
but with sides parallel to the axes is not larger than the square,
so there is no rectangle with vertices at $(0,0)$ and on the circle
whose area is larger than the chosen square.



When you do this as an exercise, I would advise using the straightforward mechanics given in some other answers rather than relying on these intuitions, under the assumption that the purpose of the exercise is to learn to use those mechanics. But you can use these intuitions to check your answer: if the $x$ and $y$ values you get are not equal, look for the mistake in your calculations.
In the simplest form...
The circle has two points where x = y along the circumference of the circle.  The largest value that x = y forms the largest area for a rectangle, if the circle resides entirely in the first quadrant and the circumference touches both the x-axis and y-axis once and only once, hence the circle's center (x,y) both x and y are both equal to the value of r. 

Largest area of a rectangle where one corner is the Origin (0,0) and the opposite corner is on the circumference of the circle.

A= ((r*(sqrt of 2)/2)+r) * ((r*(sqrt of 2)/2)+r)