Question Number 10323 by amir last updated on 04/Feb/17

Answered by mrW1 last updated on 04/Feb/17

there are 4 circles which tangent  the curve and the both coordinate axes.  they tangent the curve at point (1,1)  as well as at point(−1,−1).    let r be the radius of the circles.  (r−1)^2 +(r−1)^2 =r^2   2r^2 −4r+2=r^2   r^2 −4r+2=0  r=((4±(√(4^2 −4×2)))/2)=2±(√2)  r_1 =2−(√2)≈0.585  r_2 =2+(√2)≈3.414    r_1  is the radius of the small circles.  r_2  is the redius of the big circles.

Commented bymrW1 last updated on 04/Feb/17

Commented byamir last updated on 08/Feb/17

hello dear mrw1.thank you very much  I have another solotion for this Q.  because of the circle tangent to the   cordinate axes ,the cordinates of   its center are O ((r),(r) )  and its equation  will be (x−r)^2 +(y−r)^2 =r^(2.)   by replacing  y=(1/x) in this equation  we have   (x−r)^2 +((1/x)−r)^2 =r^2   x^2 (x−r)^2 +(1−rx)^2 =r^2 x^2   x^4 −2rx^3 +r^2 x^2 −2rx+1=0  or  x^2 r^2 −2x(x^2 +1)r+(1+x^4 )=0  r=((−b±(√(b^2 −4ac)))/(2a))=((−2x(x^2 +1)±(√(4x^2 (x^2 +1)^2 −4x^2 (1+x^4 ))))/(2x^2 ))  r=x+(1/x)±(√2)  p=2πr=2π(x+(1/x)±(√2))  (dp/dx)=0   2π(1−(1/x^2 ))=0  x^2 −1=0       x=1    x=−1  r=1+(1/1)±(√(2=))2±(√2)    ■

Commented bymrW1 last updated on 04/Feb/17

you are basically right. I have cut  short the calculation, because we  know due to symmetry that the  circles tangent the curve at (1,1)  as well as at (−1,−1).