A square with area of 4 units is partially overlapped with a circle of radius r?

You stated it a bit wrong, but the picture is clear.

Focus on just 1/8 of the circle and square in the diagram. Because of the symmetry, we only need to find the radius for which area x = area y.

The area of x = area of the sector with central angle a minus area of triangle OPQ

= πr^2(a/360) - r sin(a)/2

The area of y = area of 1/8 of the square (triangle OPR) - area of sector with central angle 45/a - area of triangle OPQ

= 1/2 - πr^2(45 - a)/360 - r sin(a)/2

Set those equal and solve.

πr^2(a/360) - r sin(a)/2 = 1/2 - πr^2(45 - a)/360 - r sin(a)/2

πr^2(a/360) = 1/2 - πr^2(45 - a)/360

(πr^2/360)(a + 45 - a) = 1/2

πr^2/8 = 1/2

πr^2 = 4

r = 2/√π = about 1.128379

i) It is a simple problem; only basing on geometry of given figure.

ii) Area of circle = Area of green shade + Area of yellow patches --------- (1)

Area of square = Area of green shade + Area of blue patches -------- (2)

But given: Area of yellow patches = Area of blue patches -------- (3)

So from (1) , (2) & (3),

Area of circle = Area of square.

iii) ==> πr² = 4

So, r² = 4/π

Solving r = 2/√π units. = 1.129 units (nearly)

My method was similar to freond... using trigonometry.

i appreciate both of the answers. the second answer, by learner required only the area of the geometric figures and solving the system of equations. great work on both parts.