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!

Guest Apr 11, 2020

#1**0 **

\(2x^2+2y^2+10x-6y-18=0\)

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

Rearrange a bit:

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

Factor

2(x^{2}+10x)+2(y^{2}-3y)=18

Divide by 2

(x^{2}+10x)+(y^{2}-3y)=9

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

(x^{2}+10x+25)+(y^{2}-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.

power27 Apr 11, 2020