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)

So,

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) ------***

So,

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.