How do you find the center and radius of the circle #x^2+y^2+6x-8y-135=0#? Precalculus Geometry of an Ellipse Identify Critical Points 1 Answer Mark D. May 2, 2018 Centre (-3,4) r=12.64911064 Explanation: #(x-a)^2+(y-b)^2=r^2# is the general formula for a circle so rearrange the equation and then complete the square. #x^2+6x+y^2-8y=135# #(x+3)^2-9+(y-4)^2-16=135# #(x+3)^2+(y-4)^2-25=135# #(x+3)^2+(y-4)^2=160# #a=-3, b=4 and r=sqrt160# 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 2803 views around the world You can reuse this answer Creative Commons License