How do you graph #x^2+y^2-1=0#? Geometry Circles Equation of a Circle 1 Answer Alan P. Dec 29, 2016 #x^2+y^2-1=0# is a circle with center at the origin and radius #1# Explanation: #x^2+y^2-1=0color(white)("XX")rArrcolor(white)("XX")x^2+y^2=1^2# Any equation of the form #x^2+y^2=r^2# is a circle with center at the origin i.e. at #(0,0)# with a radius of #r# Answer link Related questions How do you write the general form given a circle that passes through the given points. P(-2, 0),... What is the equation of the circle with a center at #(7 ,1 )# and a radius of #2 #? What is the equation of the circle with a center at #(3 ,1 )# and a radius of #2 #? What is the equation of the circle with a center at #(3 ,1 )# and a radius of #1 #? What is the equation of the circle with a center at #(3 ,5 )# and a radius of #1 #? What is the equation of the circle with a center at #(3 ,5 )# and a radius of #6 #? What is the equation of the circle with a center at #(2 ,5 )# and a radius of #6 #? What is the equation of the circle with a center at #(2 ,2 )# and a radius of #4 #? What is the equation of the circle with a center at #(2 ,2 )# and a radius of #3 #? What is the equation of the circle with a center at #(2 ,1 )# and a radius of #3 #? See all questions in Equation of a Circle Impact of this question 10930 views around the world You can reuse this answer Creative Commons License