Question Number 102060 by ajfour last updated on 06/Jul/20
Commented byajfour last updated on 06/Jul/20
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))))]](Q102140.png)
Commented bymr W last updated on 06/Jul/20
Commented byajfour last updated on 07/Jul/20
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)))) .](Q102180.png)