How do you find the center and radius of the circle #2(x-3)²+2y²=8#? Precalculus Geometry of an Ellipse Identify Critical Points 1 Answer Konstantinos Michailidis Aug 12, 2018 Divide by #2# both sides of the equation to get #(x-3)^2+y^2=4# or #(x-3)^2+(y-0)^2=2^2# Hence the center is the point #(3,0)# and the radius is #r=2# Answer link Related questions How do I find the points on the ellipse #4x^2 + y^2 = 4# that are furthest from #(1, 0)#? What are the foci of an ellipse? What are the vertices of #9x^2 + 16y^2 = 144#? What are the vertices of the graph given by the equation #(x+6)^2/4 = 1#? What are the vertices and foci of the ellipse #9x^2-18x+4y^2=27#? What are the foci of the ellipse #x^2/49+y^2/64=1#? How do I find the foci of an ellipse if its equation is #x^2/16+y^2/36=1#? How do I find the foci of an ellipse if its equation is #x^2/16+y^2/9=1#? How do I find the foci of an ellipse if its equation is #x^2/36+y^2/64=1#? How do you find the critical points for #(9x^2)/25 + (4y^2)/25 = 1#? See all questions in Identify Critical Points Impact of this question 4607 views around the world You can reuse this answer Creative Commons License