Largest rectangular area formed by origin and a point on a circle

Question

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?

Answer 1

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.

Answer 2

You are trying to maximize the equation $xy$ (the area of the rectangle) given the constraint $(x-2)^2 + (y-2)^2 = 4$. Perhaps try to isolate one of the variables in the constraint equation and use it as a substitution into the area formula? Once this is done, you could use calculus to find the maximum value.

Answer 3

Express area as a function of $x$: $A_1(x)=x \cdot (2+\sqrt{4-(x-2)^2})$. Now find the maximum of $A_1(x)$ on interval $[0,4]$. This is the case for all points of the circle where $y\ge 2$. For points where $y\le 2$, find maximum of $A_2(x)=x \cdot (2-\sqrt{4-(x-2)^2})$. Another approach is to set $x=y$ and find such points on the circle. One of them will give you the maximum area.

Answer 4

Create an equation for the area of such a rectangle and differentiate. The key is that once you select either the $x$ or the $y$ coordinate, the other is fixed because it must lie on the circle. So the equation you want to maximize is

$$A(x) = xy = x(\sqrt{4-(x-2)^2}+2)$$

This would probably be easier if you change coordinates, but we don't have to. If you differentiate with respect to $x$,

$$A'(x) = \frac{2(3x-x^2+\sqrt{-(x-4)x})}{\sqrt{-(x-4)x}}$$

Then you set the numerator equal to zero and solve, making sure the denominator also isn't equal to zero and that $-(x-4)x$ isn't negative. But this only occurs is $(x-4)x$ is positive. But based on your circle this isn't an issue because we will always have $0 \leq x \leq 4$. To ensure the denominator doesn't vanish we need $x$ with $0 < x < 4$.

There are two solutions to $A'(x) = 0$, $x = 0$ and $x = 2 + \sqrt{2}$. Only one satisfies our criterion, so the area is maximized when $x = 2 + \sqrt{2}$ and $y = 2 + \sqrt{2}$.

You can check this is a max either by the second derivative test or by checking the derivative around it. This is always a good idea. But this also confirms geometric intuition, because you might guess that a square maximizes area.

Answer 5

Since $xy=a$ is a hyperbola with the coordinate axes as asymptotes, so symmetric wrt $y=x$, and the maximum $a$ corresponds to the hyperbola tangent to the circle, which is also symmetric wrt the diagonal $y=x$, then ...

Answer 6

Maximize $z=xy$ subject to $(x-2)^2+(y-2)^2=4$.

The feasible curve (the circle) and the level curves $y=\frac{z}{x}$:

$\hspace{5cm}$

The optimal point is at $x=y$: $$(x-2)^2+(x-2)^2=4 \Rightarrow x=2+\sqrt{2}=y.$$

The maximum area is $z(2+\sqrt{2},2+\sqrt{2})=(2+\sqrt{2})^2=6+4\sqrt{2}.$

Answer 7

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)