Question Number 101375 by ajfour last updated on 02/Jul/20

Commented byajfour last updated on 02/Jul/20

In terms of R, find radii a,b,c.

Commented byajfour last updated on 05/Jul/20

Commented byajfour last updated on 06/Jul/20

lets consider both circles of radii a and c generally. Both are at a distance of (R+r) from center M of circle of radius R. Both touch the line y=R^2 +(1/4)+R both touch the parabola at (x,x^2 ) x_1 =(√(R^2 −(1/4))) center of our general circle is (2(√(Rr)) , R^2 +(1/4)+R−r) eq. of the circle: { (((x−2(√(Rr)))^2 +[x^2 −(R^2 +(1/4)+R−r)]^2 =r^2 )),(((R+(1/2))^2 −r−(r/(√(1+4x^2 ))) = x^2 )) :} and x−2(√(Rr)) = ((r(2x))/(√(1+4x^2 ))) for r=c but 2(√(Rr))−x=((r(2x))/(√(1+4x^2 ))) for r=a And since we know x_1 =(√(R^2 −(1/4))) 2(√(Rr_1 ))=x+((2x_1 r)/(√(1+4x_1 ^2 ))) 2(√(Ra))=(√(R^2 −(1/4))) + ((2a(√(R^2 −(1/4))))/(√(1+4R^2 −1))) 2(√(Ra))=(√(R^2 −(1/4))) + (a/R)(√(R^2 −(1/4))) (a/R)(√(R^2 −(1/4))) −2R(√(a/R)) +(√(R^2 −(1/4)))=0 (√(a/R))=(R/(√(R^2 −(1/4))))±(√((R^2 /(R^2 −(1/4)))−1)) (√(a/R)) =((R±(1/2))/(√(R^2 −(1/4)))) ⇒ (a/R)=(((R±(1/2))^2 )/((R+(1/2))(R−(1/2)))) It seems (a/R)=(((R+(1/2)))/((R−(1/2)))) example R=2 ⇒ a=((10)/3) ..........

Commented byajfour last updated on 05/Jul/20

Answered by mr W last updated on 05/Jul/20

Center of circle with radius R is M(0,m) y=x^2 x^2 +(y−m)^2 =R^2 ⇒y+(y−m)^2 =R^2 ⇒y^2 −(2m−1)y+m^2 −R^2 =0 Δ=(2m−1)^2 −4(m^2 −R^2 )=0 1+4R^2 −4m=0 ⇒m=(1/4)+R^2 Center of circle with radius b is B(0,m+R+b) x^2 +(y−m−R−b)^2 =b^2 y+[y−((1/4)+R^2 +R+b)]^2 =b^2 y^2 −2(R^2 +R+b−(1/4))y+(R^2 +R+b+(1/4))^2 −b^2 =0 Δ=(R^2 +R+b−(1/4))^2 −(R^2 +R+b+(1/4))^2 +b^2 =0 b^2 −b−(R^2 +R)=0 ⇒b=(1/2)(1+2R+1)=R+1 Center of circle with radius a is A(2(√(Ra)),m+R−a) it touches the parabola at P(p, p^2 ). tan θ=2p=((2(√(Ra)))/(a−R)) ⇒p=((√(Ra))/(a−R)) (2(√(Ra))−p)^2 +(m+R−a−p^2 )^2 =a^2 (2(√(Ra))−((√(Ra))/(a−R)))^2 +[m+R−a−((Ra)/((a−R)^2 ))]^2 =a^2 [2−(1/(R((a/R)−1)))]^2 ((a/R))+(1/R^2 )[m−R((a/R)−1)−((a/R)/(((a/R)−1)^2 ))]^2 =((a/R))^2 [2−(1/(R(ξ−1)))]^2 ξ+(1/R^2 )[(1/4)+R^2 −R(ξ−1)−(ξ/((ξ−1)^2 ))]^2 =ξ^2 ⇒ξ=... Center of circle with radius c is C(2(√(Rc)),m+R−c) it touches the parabola at Q(q, q^2 ). tan ϕ=2q q=2(√(Rc))+c sin ϕ=2(√(Rc))+((2cq)/(√(1+4q^2 ))) ⇒q−2(√(Rc))=((2cq)/(√(1+4q^2 ))) ...(i) q^2 =m+R−c−c cos ϕ=m+R−c−(c/(√(1+4q^2 ))) ⇒m+R−c−q^2 =(c/(√(1+4q^2 ))) ...(ii) q−2(√(Rc))=2q(m+R−c−q^2 ) q^3 −(R^2 +R−c−(1/4))q−(√(Rc))=0 ⇒q=((((√(Rc))/2)+(√(((Rc)/4)−(((R^2 +R−c−(1/4))^3 )/(27))))))^(1/3) +((((√(Rc))/2)−(√(((Rc)/4)−(((R^2 +R−c−(1/4))^3 )/(27))))))^(1/3) put this into (i) or (ii) to get c

Commented bymr W last updated on 05/Jul/20

Commented byajfour last updated on 05/Jul/20

but c is to be determined, Sir.

Commented bymr W last updated on 05/Jul/20

a and c can only be determined numerically.

Commented byajfour last updated on 05/Jul/20

if we discuss, i think we can get the expression for radii a and c , Sir..

Commented bymr W last updated on 05/Jul/20

i found no way to get them.

Commented byajfour last updated on 06/Jul/20

Sir i got a=R(R+1/2)/(R−1/2) Expression for c indeed seems difficult.

Commented bymr W last updated on 06/Jul/20

you got it at least for a! it′s correct.