It should be a square.
Draw the figure on paper, as I cannot show it here.
Let x = width of the rectangle
And y = the length
Draw any diagonal. This diagonal is a diameter of the circle so it is 2r long.
Area of rectangle, A = xy
By Pythagorean theorem, x^2 +y^2 = (2r)^2
x^2 +y^2 = 4r^2
y^2 = 4r^2 -x^2
y = sqrt(4r^2 -x^2)
A = x*sqrt(4r^2 -x^2)
Differentiate both sides with respect to x, (r is a constant),
dA/dx = x[-2x / 2sqrt(4r^2 -x^2)] +sqrt(4r^2 -x^2)
Set that to zero,
0 = x(-x) +[sqrt(4r^2 -x^2)]^2
0 = -x^2 +4r^2 -x^2
2x^2 = 4r^2
x^2 = 2r^2
x = r*sqrt(2) ------***
y = sqrt[4r^2 -2r^2] = sqrt[2r^2] = r*sqrt(2) also.
Therefore, the largest area of a rectangle in a circle whose radius is 4 is
A = 2sqrt(2) *2sqrt(2) = 4*2 = 8 sq.units ------------answer.