SOLUTION: Hi can you please help me with this question, graph the equation equal to:2x^2+2y^2-3x-5y+3=0 and find the center and radius. Thank you all your help...

Question 409765: Hi can you please help me with this question, graph the equation equal to:2x^2+2y^2-3x-5y+3=0 and find the center and radius. Thank you all your help...
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Hi can you please help me with this question, graph the equation equal to:2x^2+2y^2-3x-5y+3=0 and find the center and radius. Thank you all your help...
..
2x^2+2y^2-3x-5y+3=0
2x^2-3x+2y^2-5y=-3
completing the square
2(x^2-(3/2)x+9/16)+2(y^2-5y+25/4)=-3+9/8+25/2
2(x-3/4)^2+2(y-5/2)^2=-24/8+9/8+100/8=85/8
This is a circle with radius sqrt(85/8)=3.25, with center at(3/4,5/2)
see the graph of the circle below:
..
y=2.5+((10.625-2(x-.75)^2)/2)^.5
..

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