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]
