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In terms of π, what is the area of the circle defined by the equation 2x^2+2y^2+10x-6y-18=0?

 

This is not a homework problem. This is just for fun! 

 Apr 11, 2020
 #1
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\(2x^2+2y^2+10x-6y-18=0\)

First, you want to put this equation in a form that you can get information about the circle from.

Rearrange a bit:

2x2+10x+2y2-6y=18

Factor

2(x2+10x)+2(y2-3y)=18

Divide by 2

(x2+10x)+(y2-3y)=9

Add numbers to turn into perfect squares, making sure to add the same to both sides

(x2+10x+25)+(y2-3y+2.25)=9+25+1.5

Factor

(x+5)2+(y-1.5)2=36.25

(-5,1.5) is the center and 36.25 is the radius squared, so 36.25pi is the area.

 Apr 11, 2020
 #2
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2x^2+2y^2+10x-6y-18=0

 

2x^2+10x+2y^2-6y=18

 

2(x^2+5x)+2(y^2-3y)=18

 

(x^2+5x)+(y^2-3y)=9

 

(x^2+5x+6.25)+(y^2-3y+2.25)=9+6.25+2.25

 

 

(x+2.5)^2+(y-1.5)^2=17.5

 

17.5pi?

 Apr 11, 2020

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