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?

Guest Jan 17, 2022

#1**+8 **

The area of the square is 80.

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

Glowlife Jan 18, 2022

#2**+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\\

Melody Jan 18, 2022