What is the standard form of 4x^2+4y^2-4x+4y-8=0 of this circle?

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Answer 1 · 2018-06-23T09:13:45

#(x- 1/2)^2 + (y + 1/2)^2 = (sqrt(3/2))^2# is the standard form with #(1/2, -1/2)# the center and #sqrt(3/2)# the radius.

The standard form equation of a circle is a way to express the definition of a circle on the coordinate plane.

On the coordinate plane, the formula becomes #(x−h)^2+(y−k)^2=r^2#

h and k are the x and y coordinates of the center of the circle

#4x^2 + 4y^2 -4x + 4y -8 = 0#

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

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

#(x- 1/2)^2 + (y + 1/2)^2 = (sqrt(3/2))^2# is the standard form with #(1/2, -1/2)# the center and #sqrt(3/2)# the radius.