| Circle radius |
| X Y coordinates of the intersection of two perpendiculars, separated by a space |
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| Figure area |
Imagine a problem, there is a circle with a radius and there are two perpendicular lines whose intersection point is inside the circle.
They divide the circle into four parts. Question: what are their areas equal to?
Let's simplify the question and try to find out what the area of the shaded figure is equal to?
With due ingenuity and a little thought, this problem can be solved by anyone who knows how to calculate the segment and the area of the circle.
Let's "twist" the image in such a way that the perpendicular lines are parallel to the ordinate and abscissa axes. The area we want to find has not changed from this turn in any way.
This is what we will consider the initial data.
Shall we start counting?
\(A\) is the first circular segment whose area we can find.
\(B\) is the second circular segment, which is also easily calculated
\(S\) - this "piece" is included in both the segment \(A\) and the segment \(B\), so it is painted over with brown-green paint.
The area of the pink rectangle \(C\) is calculated as the product of \(x_0*y_0\)
The area shaded in gray on the left side of the circle is calculated as the difference between the area of a quarter of the circle and half of the segment \(A\)
And the area marked with red strokes is calculated as the difference between the area of a quarter of a circle and half of a segment \(B\)
And there was a white, unpainted area equal to a quarter of the area of the circle.
Sum it up?
\(A+B-S+C+\cfrac{R^2\pi}{4}-\cfrac{A}{2}+\cfrac{R^2\pi}{4}-\cfrac{B}{2}+\cfrac{R^2\pi}{4}=R^2\pi\)
Group, transfer and get
\(S=C+\cfrac{A}{2}+\cfrac{B}{2}-\cfrac{R^2\pi}{4}\)
Now we recall the formulas for the area of a circular segment and a rectangle.
\(C=x_0*y_0\)
\(A=\cfrac{R^2}{2}(2acos(\cfrac{y_0}{R})-sin(2acos(\cfrac{y_0}{R})))\)
\(B=\cfrac{R^2}{2}(2acos(\cfrac{x_0}{R})-sin(2acos(\cfrac{x_0}{R})))\)
In principle, it is possible to stop there, but we will go deeper in our research
Let's substitute everything into the final formula and get such a "monster"
\(x_0*y_0+\cfrac{R^2}{2}acos(\cfrac{1}{R^2}(x_0*y_0-\sqrt{(R^2-x_0^2)(R^2-y_0^2)}))-\cfrac{x_0}{2}\sqrt{R^2-x_0^2}-\cfrac{y_0}{2}\sqrt{R^2-y_0^2}=S\)
Successful calculations!