Show that the equation represents a circle by rewriting it in standard form, and find the center and radius

x^2 + y^2− 1/4x + 1/4y = 1/32
....How do i rewrite this in standard form?

Standard form of a circle is (x-a)^2 + (y-b)^2 = r^2 where (a,b) is the center and r is the radius

You'll have to complete the square twice to get it in that form

x^2 + y^2− 1/4x + 1/4y = 1/32

First, rewrite it to group the x terms together and the y terms together

x^2 − 1/4x + y^2 + 1/4y = 1/32

then add (-1/8)^2 to both sides to complete the square for x:

x^2 - 1/4x + 1/64 + y^2 + 1/4 = 1/32 + 1/64

then add another (1/8)^2 to both sides to do the same for y:

x^2 - 1/4x + 1/64 + y^2 + 1/4 + 1/64 = 1/32 + 1/64 + 1/64

then factor and simplify:

(x-1/8)^2 + (y+1/8)^2 = 1/16

so standard form would be:

(x-1/8)^2 + (y+1/8)^2 = (1/4)^2

The center is therefore (1/8, -1/8) and radius is 1/4

set it up as (x + h)^2 + (y + k)^2 = r^2

In this case, that means completing the square, for both x and y

x^2 - (1/4)x + y^2 + (1/4) y = 1/32 --> complete the square
take the coefficient of the linear term. divide it by 2. square the result and add it to both sides


x^2 - (1/4) x + 1/64 + y^2 + (1/4)y + 1/64 = 1/32 + 1/64 + 1/64

(x - 1/8)^2 + (y + 1/8)^2 = 2/32 = 1/16

center (1/8, - 1/8) -- radius = √(1/16) = 1/4

x^2 - x/4 + (1/8)^2 + y^2 + y/4 + (1/8)^2 = 1/32 + 1/64 + 1/64
(x - 1/8)^2 + (y + 1/8)^2 = 1/16
C(1/8 , -1/8) , R = 1/4