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Question Number 61215 by naka3546 last updated on 30/May/19

Answered by ajfour last updated on 30/May/19

let bottom left corner of square (i assume so, should be so) be origin. square side a , radius r centre of circle C(h,k) Then we have h^2 +k^2 =4+r^2 ...(I) h^2 +(a−k)^2 =1+r^2 ...(II) (a−h)^2 +(a−k)^2 =9+r^2 ...(III) (a−h)^2 +k^2 = x^2 +r^2 ...(IV) (IV)+(II)−(III)−(I) gives 0=x^2 +1−9−4 ⇒ x=2(√3) .

letbottomleftcornerofsquare(iassumeso,shouldbeso)beorigin.squaresidea,radiusrcentreofcircleC(h,k)Thenwehaveh2+k2=4+r2...(I)h2+(a−k)2=1+r2...(II)(a−h)2+(a−k)2=9+r2...(III)(a−h)2+k2=x2+r2...(IV)(IV)+(II)−(III)−(I)gives0=x2+1−9−4⇒x=23.

Answered by mr W last updated on 30/May/19

let r=radius of circle (1^2 +r^2 )+(x^2 +r^2 )=(2^2 +r^2 )+(3^2 +r^2 ) ⇒x^2 =2^2 +3^2 −1^2 =12 ⇒x=(√(12))=2(√3)

letr=radiusofcircle(12+r2)+(x2+r2)=(22+r2)+(32+r2)⇒x2=22+32−12=12⇒x=12=23

Commented by mr W last updated on 30/May/19

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