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Question Number 16592 by chux last updated on 24/Jun/17
please what does the question mean by the overlapping portion of A and B.
$$\mathrm{please}\:\mathrm{what}\:\mathrm{does}\:\mathrm{the}\:\mathrm{question}\:\mathrm{mean} \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{overlapping}\:\mathrm{portion}\:\mathrm{of}\:\mathrm{A}\: \\ $$$$\mathrm{and}\:\mathrm{B}. \\ $$
Commented by chux last updated on 24/Jun/17
Answered by ajfour last updated on 24/Jun/17
Commented by ajfour last updated on 24/Jun/17
Overapping area is the area common to circles A and B. Let (h,0) be the center of circle B. sin (x°/2)=(h/2) ⇒ h=2sin (x°/2) eqn of circle A: (x+h)^2 +y^2 =1 eqn of circle B: (x−h)^2 +y^2 =1 let circle A intersect x axis at (a,0). then a+h =1 or a=1−h ...(i) overlapping area is two times the area between circle A and y-axis. upper portion of circle A has eqn. y= (√(1−(x+h)^2 )) let required area is S. S=4∫_0 ^( a) (√(1−(x+h)^2 )) dx = 4{((x+h)/2)(√(1−(x+h)^2 )) + (1/2)sin^(−1) (x+h) }∣_0 ^a = 4(((a+h)/2))(√(1−(a+h)^2 ))−((4h)/2)(√(1−h^2 )) +(4/2)sin^(−1) (a+h)−(4/2)sin^(−1) h but a+h=1 and h=2sin (x°/2) S = 0−2h(√(1−h^2 ))+π−2sin^(−1) h = π−2h(√(1−h^2 ))−2sin^(−1) h . where h= 2sin (x°/2) .
$$\:\mathrm{Overapping}\:\mathrm{area}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area} \\ $$$$\mathrm{common}\:\mathrm{to}\:\mathrm{circles}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}. \\ $$$$\mathrm{Let}\:\:\left(\mathrm{h},\mathrm{0}\right)\:\mathrm{be}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{B}. \\ $$$$\:\mathrm{sin}\:\left(\mathrm{x}°/\mathrm{2}\right)=\frac{\mathrm{h}}{\mathrm{2}}\:\:\Rightarrow\:\:\mathrm{h}=\mathrm{2sin}\:\left(\mathrm{x}°/\mathrm{2}\right) \\ $$$$\mathrm{eqn}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{A}: \\ $$$$\:\:\left(\mathrm{x}+\mathrm{h}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{eqn}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{B}: \\ $$$$\:\:\left(\mathrm{x}−\mathrm{h}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{let}\:\mathrm{circle}\:\mathrm{A}\:\mathrm{intersect}\:\mathrm{x}\:\mathrm{axis}\:\mathrm{at}\:\left(\boldsymbol{\mathrm{a}},\mathrm{0}\right). \\ $$$$\mathrm{then}\:\:\:\mathrm{a}+\mathrm{h}\:=\mathrm{1}\:\:\mathrm{or}\:\:\:\:\mathrm{a}=\mathrm{1}−\mathrm{h}\:\:…\left(\mathrm{i}\right) \\ $$$$\:\mathrm{overlapping}\:\mathrm{area}\:\mathrm{is}\:\mathrm{two}\:\mathrm{times}\:\mathrm{the} \\ $$$$\mathrm{area}\:\mathrm{between}\:\mathrm{circle}\:\mathrm{A}\:\mathrm{and}\:\mathrm{y}-\mathrm{axis}. \\ $$$$\:\mathrm{upper}\:\mathrm{portion}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{A}\:\mathrm{has}\:\mathrm{eqn}. \\ $$$$\:\mathrm{y}=\:\sqrt{\mathrm{1}−\left(\mathrm{x}+\mathrm{h}\right)^{\mathrm{2}} }\: \\ $$$$\mathrm{let}\:\mathrm{required}\:\mathrm{area}\:\mathrm{is}\:\mathrm{S}. \\ $$$$\:\mathrm{S}=\mathrm{4}\int_{\mathrm{0}} ^{\:\:\mathrm{a}} \sqrt{\mathrm{1}−\left(\mathrm{x}+\mathrm{h}\right)^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$\:\:=\:\mathrm{4}\left\{\frac{\mathrm{x}+\mathrm{h}}{\mathrm{2}}\sqrt{\mathrm{1}−\left(\mathrm{x}+\mathrm{h}\right)^{\mathrm{2}} }\:+\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}+\mathrm{h}\right)\:\right\}\mid_{\mathrm{0}} ^{\mathrm{a}} \\ $$$$\:\:\:=\:\mathrm{4}\left(\frac{\mathrm{a}+\mathrm{h}}{\mathrm{2}}\right)\sqrt{\mathrm{1}−\left(\mathrm{a}+\mathrm{h}\right)^{\mathrm{2}} }−\frac{\mathrm{4h}}{\mathrm{2}}\sqrt{\mathrm{1}−\mathrm{h}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{4}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{a}+\mathrm{h}\right)−\frac{\mathrm{4}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \mathrm{h} \\ $$$$\:\mathrm{but}\:\mathrm{a}+\mathrm{h}=\mathrm{1}\:\:\mathrm{and}\:\:\mathrm{h}=\mathrm{2sin}\:\left(\mathrm{x}°/\mathrm{2}\right) \\ $$$$\:\mathrm{S}\:=\:\mathrm{0}−\mathrm{2h}\sqrt{\mathrm{1}−\mathrm{h}^{\mathrm{2}} }+\pi−\mathrm{2sin}^{−\mathrm{1}} \mathrm{h} \\ $$$$\:\:\:\:=\:\pi−\mathrm{2h}\sqrt{\mathrm{1}−\mathrm{h}^{\mathrm{2}} }−\mathrm{2sin}^{−\mathrm{1}} \mathrm{h}\:. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{where}\:\mathrm{h}=\:\mathrm{2sin}\:\left(\mathrm{x}°/\mathrm{2}\right)\:. \\ $$
Commented by chux last updated on 24/Jun/17
thanks alot sir.
$$\mathrm{thanks}\:\mathrm{alot}\:\mathrm{sir}. \\ $$
Commented by chux last updated on 24/Jun/17
please is it possible to get the value of x.
$$\mathrm{please}\:\mathrm{is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{x}. \\ $$
Commented by ajfour last updated on 24/Jun/17
that is to be chosen, in question it said 0°≤x≤60° , had it said x=30° or any other value, area of overlap shall be obtained then from the derived formula, or if the area of overlap be given the angle x can be calculated; they are interdependent.
$$\mathrm{that}\:\mathrm{is}\:\mathrm{to}\:\mathrm{be}\:\mathrm{chosen},\:\mathrm{in}\:\mathrm{question} \\ $$$$\mathrm{it}\:\mathrm{said}\:\mathrm{0}°\leqslant\mathrm{x}\leqslant\mathrm{60}°\:\:,\:\mathrm{had}\:\mathrm{it}\:\mathrm{said}\: \\ $$$$\:\mathrm{x}=\mathrm{30}°\:\mathrm{or}\:\mathrm{any}\:\mathrm{other}\:\mathrm{value},\:\mathrm{area} \\ $$$$\mathrm{of}\:\mathrm{overlap}\:\mathrm{shall}\:\mathrm{be}\:\mathrm{obtained}\:\mathrm{then} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{derived}\:\mathrm{formula},\:\mathrm{or}\:\mathrm{if}\:\mathrm{the} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{overlap}\:\mathrm{be}\:\mathrm{given}\:\mathrm{the}\:\mathrm{angle} \\ $$$$\mathrm{x}\:\mathrm{can}\:\mathrm{be}\:\mathrm{calculated};\:\mathrm{they}\:\mathrm{are} \\ $$$$\mathrm{interdependent}. \\ $$
Commented by chux last updated on 24/Jun/17
ok sir
$$\mathrm{ok}\:\mathrm{sir} \\ $$

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