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Question-46935




Question Number 46935 by ajfour last updated on 02/Nov/18
Commented by ajfour last updated on 03/Nov/18
If both coloured regions have the  same area, find locus of C in  polar form or cartesian form.  Take center of rectangle as origin,  and radius of circle R.  r=(√(x_C ^2 +y_C ^2 )) ,  tan θ = (y_C /x_C ) .
Ifbothcolouredregionshavethesamearea,findlocusofCinpolarformorcartesianform.Takecenterofrectangleasorigin,andradiusofcircleR.r=xC2+yC2,tanθ=yCxC.
Answered by ajfour last updated on 04/Nov/18
let center of circle be origin.  Eq. is   x=Rcos φ ; y = Rsin φ  center of rectangle (−rcos θ, −rsin θ)      φ= π+θ         ...(i)  upper edge:  y=−rsin θ+(b/2)  right vertical edge:      x=(a/2)−rcos θ  Let points of intersection be  A(left) , B(right).  ⇒ Rsin φ_A =−rsin θ_A +(b/2)  using (i)  ⇒   Rsin θ_A  = rsin θ_A −(b/2)  ⇒  sin θ_A =(b/(2(r−R)))  ________________________  Rcos φ_B =(a/2)−rcos θ_B   ⇒ Rcos θ_B = rcos θ_B −(a/2)  cos θ_B = (a/(2(r−R)))  ....
letcenterofcirclebeorigin.Eq.isx=Rcosϕ;y=Rsinϕcenterofrectangle(rcosθ,rsinθ)ϕ=π+θ(i)upperedge:y=rsinθ+b2rightverticaledge:x=a2rcosθLetpointsofintersectionbeA(left),B(right).RsinϕA=rsinθA+b2using(i)RsinθA=rsinθAb2sinθA=b2(rR)________________________RcosϕB=a2rcosθBRcosθB=rcosθBa2cosθB=a2(rR).

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