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The circle   \(2x^2=-2y^2+12x-4y+20\)is inscribed inside a square which has a pair of sides parallel to the x-axis. What is the area of the square?

 Jul 7, 2021

Best Answer 

 #1
avatar+35059 
+4

2x^2 - 12x     + 2y^2 + 4y      =   20       arrange this into standard circle form  ( (x-h)^2  + (y-k)^2   = r^2 

                                                                     by completing the square for  x   and y      the    radius * 2 = square side length....the area = s * s

 

divide by 2

  x^2 - 6x        +   y^2 + 2y      = 10         now complete the square for x and y        Can you take it from here? 

 Jul 7, 2021
 #1
avatar+35059 
+4
Best Answer

2x^2 - 12x     + 2y^2 + 4y      =   20       arrange this into standard circle form  ( (x-h)^2  + (y-k)^2   = r^2 

                                                                     by completing the square for  x   and y      the    radius * 2 = square side length....the area = s * s

 

divide by 2

  x^2 - 6x        +   y^2 + 2y      = 10         now complete the square for x and y        Can you take it from here? 

ElectricPavlov Jul 7, 2021
 #2
avatar+108 
+2

yea, now I know what to do, thanks ElectricPovlov!

 Jul 7, 2021

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