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The circle 2x^2 = -2y^2 + 2x - 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?

 Jan 17, 2022
 #1
avatar+254 
+8

The area of the square is 80.

Im sorry that there is no diagram. I cant upload pictures for some reason

 Jan 18, 2022
 #2
avatar+117134 
+1

2x^2 = -2y^2 + 2x - 4y + 20

 

\(2x^2 = -2y^2 + 2x - 4y + 20\\ x^2 = -y^2 + x - 2y + 10\\ (x^2-x)\;\;\;+(y^2 +2y) = 10\\ (x^2-x+0.25)\;\;\;+(y^2 +y+1) = 10+0.25+1\\ (x-0.5)^2\;\;\;+(y+1)^2 = 11.25\\\)

 

This is a circle with a radius of  \(\sqrt{11.25}\)

What will the diameter be?

So what is the area of the square that circumscribes this circle?   Hint:  sketch it.

 

 

 

LaTex:

2x^2 = -2y^2 + 2x - 4y + 20\\ 
x^2 = -y^2 + x - 2y + 10\\
 (x^2-x)\;\;\;+(y^2 +2y) = 10\\
 (x^2-x+0.25)\;\;\;+(y^2 +y+1) = 10+0.25+1\\
 (x-0.5)^2\;\;\;+(y+1)^2 = 11.25\\

 Jan 18, 2022
edited by Melody  Jan 18, 2022

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