So I would need to solve how to define the intersection area of a circle and an ellipse if they had 4 intersection points. Radius of circle is 2000 and it's x- and y- coordinates are (5000,5000). Ellipse's x- and y-coordinates are also (5000,5000), semi-major axis = 3500 and minor-axis = 1500. 
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I would appreciate a ton if someone could help me with practical solution step by step.
I know that first we have to figure out the intersection points and after that integrate the area but I don't know how to do it in practice.
Best Answer
Put the center of ellipse and circle at the origin. The equation of ellipse is:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\tag{1}$$
...and the equation of the circle is $$x^2+y^2=r^2\tag{2}$$
...with $a=3500,b=1500,r=2000$
Solve equautions (1) and (2) for positive values of $x,y$ and you'll get coordinates of point $A$:
$$x_A=175\sqrt{70},\ \ y_A=75 \sqrt{330}$$
Now you have to calculate the area $CAB$ between curves $BA$ and $CA$ and vertical segment $BC$. The equation of curve $AB$ is:
$$y=\frac ba\sqrt{a^2-x^2}$$
...and the equation of curve $AC$ is:
$$y=\sqrt{r^2-x^2}$$
So the area of $P_{ABC}$ is:
$$P_{ABC}=\int_0^{x_A}(\sqrt{r^2-x^2}-\frac ba\sqrt{a^2-x^2})dx$$
$$P_{ABC}=125000 \left(16 \sin ^{-1}\left(\frac{7 \sqrt{\frac{7}{10}}}{8}\right)-21 \sin ^{-1}\left(\frac{\sqrt{\frac{7}{10}}}{2}\right)\right)\approx509768$$
The are of intersection is:
$$P=r^2\pi-4P_{ABC}\approx 1.05273\times10^7$$