How do you find the center and radius of the circle #2x²+2y²-5x+y=0#?

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Answer 1 · 2017-02-11T17:28:05

This is the equation of a circle, center #(5/4,-1/4)# and radius #sqrt(13/8)#

Let's rearrange the equation and complete the squares

#2x^2-5x+2y^2+y=0#

#2(x^2-5/2x)+2(y^2+1/2y)=0#

#2(x^2-5/2x+25/16)+2(y^2+1/2y+1/16)=25/8+1/8#

#2(x-5/4)^2+2(y+1/4)^2=26/8#

#(x-5/4)^2+(y+1/4)^2=26/16=13/8#

This is the equation of a circle, center #(5/4,-1/4)# and radius #sqrt(13/8)#