Question Number 102078 by Jamshidbek2311 last updated on 06/Jul/20

Answered by mr W last updated on 06/Jul/20

METHOD I  x^2 +y^2  is the square of the distance  from point (x,y) to the origin (0,0).    (x+5)^2 +(y−12)^2 =14^2  is the circle  with radius 14 and center at (−5,12).  distance from (−5,12) to the origin  is (√((−5)^2 +12^2 ))=13, the minimum  distance from a point on the circle to  the origin is then 14−13=1 and the   maximum distance is 14+13=27.  ⇒(x^2 +y^2 )_(min) =1^2 =1  ⇒(x^2 +y^2 )_(max) =27^2 =729

Answered by mr W last updated on 06/Jul/20

METHOD II  k=x^2 +y^2   F=x^2 +y^2 +λ[(x+5)^2 +(y−12)^2 −14^2 ]  (∂F/∂x)=2x+2λ(x+5)=0 ⇒x=−((5λ)/(λ+1))  (∂F/∂y)=2y+2λ(y−12)=0 ⇒y=((12λ)/(λ+1))  (−((5λ)/(λ+1))+5)^2 +(((12λ)/(λ+1))−12)^2 −14^2 =0  5^2 (1−(λ/(λ+1)))^2 +12^2 (1−(λ/(λ+1)))^2 −14^2 =0  13^2 (1−(λ/(λ+1)))^2 =14^2   ⇒1−(λ/(λ+1))=±((14)/(13))  ⇒λ+1=±((13)/(14))  ⇒λ=−(1/(14)) or −((27)/(14))  with λ=−(1/(14)):  x=−5×((−(1/(14)))/(1−(1/(14))))=(5/(13))  y=12×((−(1/(14)))/(1−(1/(14))))=−((12)/(13))  ⇒k=x^2 +y^2 =((5/(13)))^2 +(−((12)/(13)))^2 =1=min.  with λ=−((27)/(14)):  x=−5×((−((27)/(14)))/(1−((27)/(14))))=−5×((27)/(13))  y=12×((−((27)/(14)))/(1−((27)/(14))))=12×((27)/(13))  ⇒k=x^2 +y^2 =(−5×((27)/(13)))^2 +(12×((27)/(13)))^2 =729=max.

Answered by mr W last updated on 06/Jul/20

METHOD III  k=x^2 +y^2 =d^2   let x=d cos θ, y=d sin θ  (d cos θ+5)^2 +(d sin θ−12)^2 =14^2   d^2 +2d(5 cos θ−12 sin θ)−27=0  d^2 +26d(sin α cos θ−cos α sin θ)−27=0  d^2 +26d sin (α−θ)−27=0  d=−13 sin (α−θ)+(√(169 sin^2  (α−θ)+27))  d_(max) =13+(√(169+27))=13+14=27  d_(min) =−13+(√(169+27))=−13+14=1  k=x^2 +y^2 =d^2 =27^2 =729=max.  k=x^2 +y^2 =d^2 =1^2 =1=min.

Answered by mr W last updated on 06/Jul/20

METHOD IV  x=−5+14 cos θ  y=12+14 sin θ  x^2 +y^2 =5^2 +12^2 +14^2 +28(−5 cos θ+12 sin θ)  =365+28×13(−sin α cos θ+cos α sin θ)  =365+28×13 sin (θ−α)  max. =365+28×13=729  min. =365−28×13=1

Answered by john santu last updated on 07/Jul/20

method JS  x^2 +y_(min) ^2  = [ r −(√(x_c ^2 +y_c ^2 )) ] ^2   x^2 +y_(max) ^2  = [ r +(√(x_c ^2 +y_c ^2 )) ]^2