Where is the center of the graph of #(x+3)^2+(y-2)^2=4#? Precalculus Translation and Rotation of Axis Translation of a Conic Section 1 Answer AJ Speller Sep 11, 2014 Standard form: #(x-h)^2+(y-k)^2=r^2# where #(h,k)# is the center of the circle and #r# is the radius of the circle The center of the circle is #(-3,2)#. You need to use the opposite signs for the numbers used in place of #(h,k)# Answer link Related questions How do you translate a graph to the left or right? How do you translate a graph up or down? What does it mean to translate a graph? If the equation of a conic section is #(x-3)^2/49+(y-4)^2/13=1#, how has its center been translated? If the equation of a conic section is #(x+6)^2/81-(y+1)^2/16=1#, how has its center been translated? If the equation of a conic section is #(x-2)^2+(y+5)^2=25#, how has its center been translated? What is the difference between the graph of #y=2x^2+3x-4# and the graph of #y=2(x-4)^2+3(x-4)-4#? How do use the method of translation of axes to sketch the curve #x^2+y^2-4x+3=0#? How do use the method of translation of axes to sketch the curve #x^2+y^2+2x-6y+6=0#? How do use the method of translation of axes to sketch the curve #x^2-y^2+4x-2y+2=0#? See all questions in Translation of a Conic Section Impact of this question 2185 views around the world You can reuse this answer Creative Commons License