Question Number 102060 by ajfour last updated on 06/Jul/20

Commented byajfour last updated on 06/Jul/20

Find r and R.

Answered by mr W last updated on 06/Jul/20

center of circle r inside M(0,m)  m=(1/4)+r^2   center of circle r outside N((√(4r^2 −((1/4)+r^2 −r)^2 )),r)  (1/2)(r+(1/4)+r^2 )=(1/4)[4r^2 −((1/4)+r^2 −r)^2 ]  2r^2 −2r−(1/2)=((1/4)+r^2 −r)^2   r^4 −2r^3 −(r^2 /2)+((3r)/2)+(9/(16))=0  (r−(3/2))^2 (r+(1/2))^2 =0  ⇒r=(3/2)    center of circle R at (R,h)  R^2 +(h−(1/4)−r^2 )^2 =(R+r)^2    ...(i)  ((√(4r^2 −((1/4)+r^2 −r)^2 ))−R)^2 +(h−r)^2 =(R+r)^2    ...(ii)  (i)−(ii):  (2R−(√(4r^2 −((1/4)+r^2 −r)^2 )))(√(4r^2 −((1/4)+r^2 −r)^2 ))+(2h−(1/4)−r^2 −r)(−(1/4)−r^2 +r)=0  ⇒h=2((√2)R−1)  R^2 +(2(√2)R−(9/2))^2 =(R+(3/2))^2   8R^2 −3(1+6(√2))R+18=0  ⇒oR=(3/(16))[1+6(√2)+(√(3(3+4(√2))))]

Commented bymr W last updated on 06/Jul/20

Commented byajfour last updated on 07/Jul/20

Thank you Sir, Excellent!

Answered by ajfour last updated on 07/Jul/20

let the smaller circles touch parabola  at P(h,h^2 )  h=rsin θ   ,    h^2 =r+rcos θ  And     2h=tan θ  ⇒   tan θ=2rsin θ   ⇒   cos θ=(1/(2r))  &     r^2 (1−(1/(4r^2 )))=r+(1/2)  ⇒    r^2 −(1/4)=r+(1/2)          r^2 −r−(3/4)=0    ⇒  r=(1/2)+(√((1/4)+(3/4)))      r=(3/2)       ,   cos θ=(1/3)  center of left smaller circle be C.  y_C = r+2rcos θ = (3/2)+1 = (5/2)    Altitude of isosceles triangle       H=(√((R+r)^2 −r^2 ))  Hcos θ+rsin θ=R  ⇒  (1/9)(R^2 +2Rr)=(R−(3/2)×((2(√2))/3))^2   ⇒   R^2 +3R=9[R^2 −2(√2)R+2]  ⇒    8R^2 −(18(√2)+3)R+18=0     R=((18(√2)+3)/(16))+(√((((18(√2)+3)/(16)))^2 −((18)/8)))     R = (3/(16))(6(√2)+1+(√(9+12(√2)))) .