Let's call the radius of circle #1 as r. That of the circle #2 as R.
For circle #2 to overlap half of the area of circle #1, R must be grater than r.
What your question is asking for is not very specific. All it says is that circle #2 overlaps half of the area of circle #1. There are infinitely many R's to do that.
One is when R = 2r. Circle #2 will not only ovrlap half of the area of circle #1....it will cover all of the area of circle #1 too. But this r will satisfy the question because with it, circle #2 overlaps half of the area of circle #1. This R is the maximum R then.
Another R is when the intersections of the two circles are at the end of the diameter of circle #1 that is perpendicular to the center of circle #2.
Draw the figure such that the center of circle #2 is at the right end of the horizontal diameter of the circle #1...at point P. Then the circle #2 passes through the top and bottom ends of the vertical diameter of circle #1.
Draw a R from P to the top end of the vertical diameter of circle #1....point A.
Call the center of circle #1, point O.
In right triangle AOP.
AO = OP = r
AP = R
By Pythagorean theorem, R = sqrt(r^2 +r^2+ = r*sqrt(2)
Therefore R = r*sqrt(2) is another answer to your question.
But I think your question should have been more specific. It should have asked the minimum R that will overlap half of the area of circle #1, or what sets of R's that will overlap half of the area of circle #1.
Er, do I misunderstand your question as posted? Do you mean what R that will overlap exactly half of the area of circle #1?
To get the R that will make circle #2 occupy exactly half of the area of circle #1, here is one way.
Draw the figure. Almost the same in the beginning as the figure above, except that now the intersection of the two circles are to the right of the vertical diameter of circle #1.
Draw a vertical line segment, a mutual chord to both circles, joining the two intersection points. Call the top, middle and bottom parts of this chord as B,C,D respectively.
Draw a left horizontal R. It will pass through C, and O,(being the center of circle #1), and will end at point Q.
Draw R PB, and r OB.
The area of circle #1 that is being occupied by circle #2 is twice the area of the figure PCOQB.
This figure PCOQB is (A1...sector PQB of circle #2) + (A2... circular secant BP of circle #1).
Let theta, or t, be the central angle BOP.
Triangle BOP is isoscles because BO = OP = r.
So, angle OBP = angle OPB = (1/2)(2pi -t) = (pi -t/2) radians each.
A1 = (1/2)(angle BOQ)(R^2)
A1 = (1/2)(pi -t/2)(R^2)
A2 = (1/2)(t)(r^2) -(1/2)(r*r*sin(t))
A2 = (1/2)(r^2)[t -sin(t)]
Total area occupied, A = 2(area PCOQB) = 2(A1 +A2)
A = (pi -t/2)(R^2) +[t -sin(t)](r^2)
This A is also (1/2)pi(r^2), so,
(1/2)pi(r^2) = (pi -t/2)(R^2) +[t -sin(t)](r^2) -----(1)
You Know only r here. It should be a constant.
The R and the angle t are variables. So we need another independent equation involving R and t.
Back to the figure.
r = line segment PCO.
r = PC +CO
r = R*cos(pi -t/2) +r*cos(t) ---------(2)
There. Solve Eqs. (1) and (2) simultaneously to find R.
If you have time to do those, ....get set, .....fire!