so you are trying to find the area of intersection for a circle and square in the 2nd quadrant.

the top of the square is y=100, x starts at -100 and goes to 0.

instead of thinking of the region as three regions (two triangles and a wedge) think of it as two regions (area under curve and area under flat line)

the equation for a circle with radius r is x^2 + Y^2 = r^2

but since your region is in the 2nd quadrant, y is positive and thus:

y= sqrt(r^2-x^2)

to find where the curve stops, set y=100 and get

r^2-x^2=10,000

x^2 = r^2-10,000

x=sqrt(r^2-10,000)

the area of the non curvy region is (100-sqrt(r^2-10,000))*100

the area of the curvey part is: oh boy here we go

the definite integral from x=100 to x=sqrt(r^2-10,000) of sqrt(r^2-x^2) which equals (i believe) 1/r arcsin (sqrt(r^2-10,000)/r) - 1/r arcsin(100/r)

I am really high right now and do not have paper so you might want to do the calculus yourself. Add the area of the curvy and non curvey regions and set equal to 3500, then solve for r. Solving for r will such. Try using inverse funtions on both sides at first and then just resort to newtons method or something. However, rest assured that if a solution to your problem exists, it will be the solution to this equation.

again, i am stoned out of my mind so make sure the integral of sqrt(r^2-x^2) really is what I said it is. Good luck and happy equation solving.