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Question Number 71034 by ajfour last updated on 11/Oct/19
Commented by ajfour last updated on 11/Oct/19
Commented by ajfour last updated on 11/Oct/19
Commented by ajfour last updated on 11/Oct/19
Commented by mr W last updated on 12/Oct/19
![A(0,−(R−a)) B((R−b)cos α,(R−b)sin α) C((R−c)cos β,(R−c)sin β) X(p,q) with radius r (p−0)^2 +[q+(R−a)]^2 =(r+a)^2 ...(i) [p−(R−b)cos α]^2 +[q−(R−b)sin α]^2 =(r+b)^2 ...(ii) [p−(R−c)cos β]^2 +[q−(R−c)sin β]^2 =(r+c)^2 ...(iii) (i)−(ii): (R−b)cos α[2p−(R−b)cos α]+[(R−a)+(R−b)sin α][2q+(R−a)−(R−b)sin α]=(a−b)(2r+a+b) (R−c)cos β[2p−(R−c)cos β]+[(R−a)+(R−c)sin β][2q+(R−a)−(R−c)sin β]=(a−c)(2r+a+c) with A=R−a, B=R−b, C=R−c, P=2p, Q=2q Bcos α[P−Bcos α]+[A+Bsin α][Q+A−Bsin α]=(a−b)(2r+a+b) Ccos β[P−Ccos β]+[A+Csin β][Q+A−Csin β]=(a−c)(2r+a+c) Bcos αP+(A+Bsin α)Q=2(r+R)(B−A) ...(I) Ccos βP+(A+Csin β)Q=2(r+R)(C−A) ...(II) [A(B cos α−C cos β)+BC sin (β−α)]Q=2(r+R)[BC(cos a−cos β)−A(B cos α−C cos β)] ⇒Q=((2(r+R)[BC(cos a−cos β)−A(B cos α−C cos β)])/(A(B cos α−C cos β)+BC sin (β−α))) ⇒q=(r+R)(([BC(cos a−cos β)−A(B cos α−C cos β)])/(A(B cos α−C cos β)+BC sin (β−α))) ⇒q=(r+R)V [A(B cos α−C cos β)+BC sin (β−α)]P=2(r+R)[A(B−C)+BC(sin β−sin α)−A(Csin β−Bsin α)] ⇒P=((2(r+R)[A(B−C)+BC(sin β−sin α)−A(Csin β−Bsin α)])/(A(B cos α−C cos β)+BC sin (β−α))) ⇒p=(r+R)(([A(B−C)+BC(sin β−sin α)−A(Csin β−Bsin α)])/(A(B cos α−C cos β)+BC sin (β−α))) ⇒p=(r+R)U from (i): U^2 (r+R)^2 +[V(r+R)+A]^2 =(r+R−A)^2 let λ=r+R ⇒U^2 λ^2 +(Vλ+A)^2 =(λ−A)^2 ⇒(1−U^2 −V^2 )λ=2A(V+1) ⇒λ=r+R=((2A(V+1))/(1−U^2 −V^2 )) ⇒r=((2A(V+1))/(1−U^2 −V^2 ))−R=f(α,β) with A=R−a B=R−b C=R−c U=((A(B−C)+BC(sin β−sin α)−A(Csin β−Bsin α))/(A(B cos α−C cos β)+BC sin (β−α))) V=((BC(cos a−cos β)−A(B cos α−C cos β))/(A(B cos α−C cos β)+BC sin (β−α))) (∂r/∂α)=0 ⇒ eqn. 1 with α,β (∂r/∂β)=0 ⇒ eqn. 2 with α,β ...](Q71100.png)
Commented by peter frank last updated on 11/Oct/19
Commented by ajfour last updated on 12/Oct/19
