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Question-105001




Question Number 105001 by ajfour last updated on 25/Jul/20
Commented by ajfour last updated on 25/Jul/20
If both circles have unit radius, and  regions 1, 2, 3, 4, 5 have equal areas,  find eq. of both circles.
$${If}\:{both}\:{circles}\:{have}\:{unit}\:{radius},\:{and} \\ $$$${regions}\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4},\:\mathrm{5}\:{have}\:{equal}\:{areas}, \\ $$$${find}\:{eq}.\:{of}\:{both}\:{circles}. \\ $$
Answered by ajfour last updated on 25/Jul/20
let lower circle eq. be      (x−h)^2 +(y+k)^2 =1  upper circle eq. is then      (x+k)^2 +(y−h)^2 =1  let A(a,0)   & B(b,0)   let  x_0 =a, b  ⇒  x_0 =h±(√(1−k^2 ))  ⇒   a=h+(√(1−k^2 ))   ,   b=h−(√(1−k^2 ))   P (p,p) lies on y=x and both circles  hence   (p+k)^2 +(p−h)^2 =1  A_5 = ∫_b ^( 0) (−k+(√(1−(x−h)^2 )) )dx  A_2 +A_3  = ∫_0 ^(  a) (−k+(√(1−(x−h)^2 )) )dx  A_3  = 2∫_0 ^( p) (−k+(√(1−(x−h)^2 ))−x)dx  Now    A_2 +A_3  = 2A_5    ....(i)  &                 A_3  = A_5               ....(ii)  .....
$${let}\:{lower}\:{circle}\:{eq}.\:{be} \\ $$$$\:\:\:\:\left({x}−{h}\right)^{\mathrm{2}} +\left({y}+{k}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${upper}\:{circle}\:{eq}.\:{is}\:{then} \\ $$$$\:\:\:\:\left({x}+{k}\right)^{\mathrm{2}} +\left({y}−{h}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${let}\:{A}\left({a},\mathrm{0}\right)\:\:\:\&\:{B}\left({b},\mathrm{0}\right) \\ $$$$\:{let}\:\:{x}_{\mathrm{0}} ={a},\:{b} \\ $$$$\Rightarrow\:\:{x}_{\mathrm{0}} ={h}\pm\sqrt{\mathrm{1}−{k}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\:{a}={h}+\sqrt{\mathrm{1}−{k}^{\mathrm{2}} }\:\:\:,\:\:\:{b}={h}−\sqrt{\mathrm{1}−{k}^{\mathrm{2}} }\: \\ $$$${P}\:\left({p},{p}\right)\:{lies}\:{on}\:{y}={x}\:{and}\:{both}\:{circles} \\ $$$${hence}\:\:\:\left({p}+{k}\right)^{\mathrm{2}} +\left({p}−{h}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${A}_{\mathrm{5}} =\:\int_{{b}} ^{\:\mathrm{0}} \left(−{k}+\sqrt{\mathrm{1}−\left({x}−{h}\right)^{\mathrm{2}} }\:\right){dx} \\ $$$${A}_{\mathrm{2}} +{A}_{\mathrm{3}} \:=\:\int_{\mathrm{0}} ^{\:\:{a}} \left(−{k}+\sqrt{\mathrm{1}−\left({x}−{h}\right)^{\mathrm{2}} }\:\right){dx} \\ $$$${A}_{\mathrm{3}} \:=\:\mathrm{2}\int_{\mathrm{0}} ^{\:{p}} \left(−{k}+\sqrt{\mathrm{1}−\left({x}−{h}\right)^{\mathrm{2}} }−{x}\right){dx} \\ $$$${Now}\:\:\:\:{A}_{\mathrm{2}} +{A}_{\mathrm{3}} \:=\:\mathrm{2}{A}_{\mathrm{5}} \:\:\:….\left({i}\right)\:\:\& \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{A}_{\mathrm{3}} \:=\:{A}_{\mathrm{5}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({ii}\right) \\ $$$$….. \\ $$

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