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).