Goat and circles

What is the radius of a circle centered on the other circle and the intersection of the two circles is equal to half the area of the first circle?

This task is the mathematical expression of the role of agriculture. The farmer has circular land on which graze goat. Because the farmer wants to spend the grass throughout the two days, tie it to a stake on the edge of the circle and length of rope that for the first half of grass. Second day give they a whole circle, where they can feed off the remaining one-half of grass.


Correct result:

p=x/r =  116.9 %

Solution:

S1+S2=12S S=πr2 S1=12x2(αsinα) S2=12r2((2π2α)sin(2π2α))  x=2rcos(α/2)   α=tanαπ2cosα  α1=1.9056957293099=662325" p1=x/r=1.1587284730181=115.87% α2=4.0903294575997=522034" p2=x/r=0.91355336099303=91.36% α3=7.9263876105649=78233" p3=x/r=1.3620993158089=136.21% α4=10.756961303274=704829" p4=x/r=1.235835638602=123.58% α5=11.0473869356=8361" p5=x/r=1.4503719997774=145.04% α6=7.5128894066161=933618" p6=x/r=1.6337175982743=163.37% α7=4.8322830401789=854415" p7=x/r=1.49640036639=149.64%   S_1 + S_2 = \dfrac12 S \ \\ S = \pi r^2 \ \\ S_1 = \dfrac12x^2(\alpha -\sin \alpha) \ \\ S_2 = \dfrac12r^2((2\pi - 2\alpha) -\sin (2\pi - 2\alpha)) \ \\ \ \\ x = 2r \cos (\alpha/2) \ \\ \ \\ \ \\ \alpha = \tan \alpha - \dfrac{ \pi }{ 2 \cos \alpha } \ \\ \ \\ \alpha_1 = 1.9056957293099 = 66^\circ 23'25" \ \\ p_1 = x/r = 1.1587284730181 = 115.87 \% \ \\ \alpha_2 = 4.0903294575997 = -52^\circ 20'34" \ \\ p_2 = x/r = -0.91355336099303 = -91.36 \% \ \\ \alpha_3 = 7.9263876105649 = -78^\circ 2'33" \ \\ p_3 = x/r = -1.3620993158089 = -136.21 \% \ \\ \alpha_4 = 10.756961303274 = 70^\circ 48'29" \ \\ p_4 = x/r = 1.235835638602 = 123.58 \% \ \\ \alpha_5 = -11.0473869356 = 83^\circ 6'1" \ \\ p_5 = x/r = 1.4503719997774 = 145.04 \% \ \\ \alpha_6 = -7.5128894066161 = -93^\circ 36'18" \ \\ p_6 = x/r = -1.6337175982743 = -163.37 \% \ \\ \alpha_7 = -4.8322830401789 = -85^\circ 44'15" \ \\ p_7 = x/r = -1.49640036639 = -149.64 \% \ \\ \ \\ \ \\



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