I'm searching a Radius R of a circle, which bisects the area of a given circle with radius r. The center of the searched circle ( radius R) lies on the circumference of the given circle ( radius r). I've problems with the integration limits arcsin(R/2r) and pi-arcsin(R/2r) of the solve function. Has anyone an idea, how to do this with mathematica ? Thank you so much !

In[24]:= Part[Reduce[2 r Sin[x] == R, x], 2]

Out[24]= C[1] \[Element] Integers &&

r != 0 && (x == \[Pi] - ArcSin[R/(2 r)] + 2 \[Pi] C[1] ||

x == ArcSin[R/(2 r)] + 2 \[Pi] C[1])

Solve[2 r Sin[x] == R, x]

{{x -> ConditionalExpression[\[Pi] - ArcSin[R/(2 r)] + 2 \[Pi] C[1],

C[1] \[Element] Integers]}, {x ->

ConditionalExpression[ArcSin[R/(2 r)] + 2 \[Pi] C[1],

C[1] \[Element] Integers]}}

In[19]:= Part[{{x ->

ConditionalExpression[\[Pi] - ArcSin[R/(2 r)] + 2 \[Pi] C[1],

C[1] \[Element] Integers]}, {x ->

ConditionalExpression[ArcSin[R/(2 r)] + 2 \[Pi] C[1],

C[1] \[Element] Integers]}}, 1]

Out[19]= {x ->

ConditionalExpression[\[Pi] - ArcSin[R/(2 r)] + 2 \[Pi] C[1],

C[1] \[Element] Integers]}

Integrate[x, {x, R, 2 r Sin[s]}]

-(R^2/2) + 2 r^2 Sin[s]^2

In[22]:= Solve[

Integrate[

x, {x, R,

2 r Sin[s]}, {s, Reduce[2 r Sin[x] == R, x],

Reduce[2 r Sin[x] == R, x]} ] == Pi/2 r r, R]