We are tasked with expressing the area $A(x)$ of a rectangle inscribed in a circle of radius 6 as a function of $x$, where $P = (x, y)$ is a vertex of the rectangle.

Step-by-Step Breakdown:

1. Equation of the Circle:
   The equation of a circle with center at the origin and radius $6$ is: $x^2 + y^2 = 36$ This equation will help us relate $x$ and $y$ for any point $P(x, y)$ on the circle.

2. Dimensions of the Rectangle:
   The rectangle is inscribed in the circle. Given the symmetry of the figure, if $P(x, y)$ is a vertex of the rectangle in Quadrant I, then the opposite vertex of the rectangle would be at $(-x, -y)$.

   • The width of the rectangle (along the $x$-axis) would be $2x$.
   • The height of the rectangle (along the $y$-axis) would be $2y$.

3. Area of the Rectangle:
   The area $A(x)$ of the rectangle is the product of its width and height: $A(x) = \text{width} \times \text{height} = 2x \times 2y = 4xy$

4. Substitute $y$ in Terms of $x$:
   From the equation of the circle, we can solve for $y$ in terms of $x$: $y = \sqrt{36 - x^2}$ Thus, the area function becomes: $A(x) = 4x \cdot \sqrt{36 - x^2}$

Final Answer:

$A(x) = 4x \sqrt{36 - x^2}$

This is the expression for the area of the rectangle as a function of $x$.

Would you like to explore any further details on this? Here are five related questions:

1. How can we determine the maximum possible area of the rectangle?
2. What is the domain of the function $A(x)$?
3. How does the graph of $A(x)$ look, and where does it reach its maximum?
4. Can we solve for $x$ when the area is a given value?
5. How can we extend this problem to a general circle of radius $r$?

Tip: Always double-check the domain of your function to ensure it makes sense in the context of the problem. For example, $x$ must be between $-6$ and $6$, inclusive, because the radius of the circle is 6.