Question Number 103756 by ajfour last updated on 17/Jul/20

Commented byajfour last updated on 17/Jul/20

Find radius r of shown circle.

Answered by mr W last updated on 17/Jul/20

center of circle (h,r)    y=x^2   tan θ=y′=2x  P(p,p^2 )  p=h−r sin θ  p^2 =r+r cos θ  tan θ=2p  p=h−r ((2p)/(√(1+4p^2 )))  p^2 =r(1+ (1/(√(1+4p^2 ))))   { ((h=p(1+((2p^2 )/(1+(√(1+4p^2 ))))))),((r=(p^2 /(1+(1/(√(1+4p^2 ))))))) :}     ...(I)    y=(√x)  tan ϕ=y′=(1/(2(√x)))  Q(q^2 ,q)  q^2 =h−r sin ϕ  q=r+r cos ϕ  tan ϕ=(1/(2q))  q^2 =h−r (1/(√(1+4q^2 )))  q=r(1+((2q)/(√(1+4q^2 ))))   { ((h=q(q+(1/(2q+(√(1+4q^2 ))))))),((r=(q/(1+((2q)/(√(1+4q^2 ))))))) :}   ...(II)  intersection point of curve (I) & (II):  h≈1.4167  r≈0.56886

Commented bymr W last updated on 17/Jul/20

Commented bymr W last updated on 17/Jul/20

Commented byajfour last updated on 17/Jul/20

parametric way!  Anyway thanks  Sir,  i can barely think of any  other way out...