excuse my basic knowledge of maths :(
So I have a square and a circle which is overlapping it and I would wish to find the area of this shape and split this with a line so that the area of this shape would be equal.
Many thanks, I'd be grateful for the support :)
On
Find the intersection points of the (boundaries of the) circle and the square. Let's assume there is one corner of the square inside the circle like in the picture, so that the bounding path consists of a piece of the circle and two line segments which are parts of sides of the square.
Let's try to use Green's Theorem to find the area $A$ of the intersection. Parametrice the circle as $r(\cos(t), \sin(t)), \space t\in[0, 2\pi]$. Let the intersection points be with $t=a$ (the above one) and $t=b$ (the lower one).
We get
$$A = \int_{\partial A} x dy = \int_{a}^b r^2\cos^2t \space dt + 0 + \int_{r \sin b}^{r\sin a} r \cos a \space dt$$
$$=r^2 \left ( \frac{b-a}{2} + \sin a \cos a - \sin b \cos b \right )$$ $$ = \frac{r^2}{2} \left ( b-a + \sin(2a) - \sin(2b) \right)$$
PS. I'm little bit uncertain if this is true or if I made an error. I'd be very happy if someone would check it. Also, sorry for the questioner if this is too evolved a method (since he/she mentioned basic knowledge) but I thought it might be a cool way of calculating the area.
The area of a circular segment is $$A_1=\frac{R^2}{2}(\theta-\sin(\theta))$$ (cf. https://en.wikipedia.org/wiki/Circular_segment#Area)
The area of the spaning triangle simply is $$A_2=\frac{R^2}{2}\sin(\theta)$$ (alternatively: cf. https://en.wikipedia.org/wiki/Circular_sector#Area)
Thus your question about the depth of overlap amounts in solving for $$0=A_1-A_2=\frac{R^2}{2}(\theta-2\sin(\theta))$$
This then numerically results in $\theta=1.895494267$ (in radians) or a height of the latter triangle of $$h/R=\cos(\theta/2)=0.583514128$$ and the asked for depth of overlap thus becomes finally $$\frac{R-h}{R}=0.416485872$$
--- rk