Find the largest real x number for which there exists a real number y such that x^2 + y^2 = 2x + 2y.

x^2 + y^2  =  2x + 2y

 

x^2  - 2x  + y^2 - 2y  = 0

 

Complete the square on x and y  and we have that

 

x^2 - 2x + 1   +  y^2 -2y + 1  =  2

 

( x - 1)^2  +  (y - 1)^2  =  2

 

x will be maximized when  y = 1...so....

 

(x - 1)^2  =  2       take the positive square root

 

x - 1  =  √2

 

x =  √2 + 1

 

 

  

CPhill Nov 6, 2017